Answers:
(a) The explanatory variable is the change in the temperature .
The response variable is the coral survivance.
(b) The explanatory variable is quantitative.
The response variable is categorical.
Explanation:
A quantitative variable is those that can take a range of values. In this case, the temperature may vary across a range of values associated with the survival of coral reefs.
A categorical (also known as qualitative) variable is characterized to present finite states. In this case, the coral reefs can survive or not depending on their responses to changes in the temperature.
Response and explanatory variables are used in experimental procedure ; the explanatory and response variables are :
Measured variable - > Change in weight of coral (quantitative) Explanatory variable - > Temperature of water. (categorical)Explanatory variables are independent variables which causes changes to the measured variable.
Measured variable is the variable we intend to measure in an experiment. They are called dependent variables.
Quantitative variables are numerical values, weight is measured as a number. Hence, it is quantitative.
Categorical variables are labels and are used to group objects into classes. They are Non-numeric. Levels of temperature could be (high, medium, low). Hence, temperature is categorical in the experiment.
Learn more :https://brainly.com/question/17498238?referrer=searchResults
Interpret and describe an example of a chemical formula. Summarize the two parts a chemical formula must have.
Answer and Explanation:
A chemical formula is composed by chemical symbols (in letters, which indicate the chemical elements) and subscripts (numbers). For example, let see the chemical formula for carbon dioxide:
CO₂
The letters C and O are the chemical symbols for the elements carbon (C) and oxygen (O). The number 2 in subscript indicates that there are 2 atoms of O per molecule. The subscript 1 is generally not indicated in chemical formulae, so it is assumed that the number of atoms of C is 1.
Summarizing, the chemical formula CO₂ indicates us that the molecule if formed by 1 atom of the element carbon (C) and 2 atoms of the element oxygen (O).
Given the reaction: 2Na(s) + 2H20(1) 2Na+(aq) + 2OH(aq) + H2(g)
This reaction goes to completion because one of the products formed is
1.
an insoluble base
2.
a soluble base
3
a precipitate
4.
a gas
A soluble base is formed when sodium reacts with water.
What happens when sodium reacts with water?
When sodium reacts with water, it produces strongly alkalic sodium hydroxide which is also called caustic soda and hydrogen gas. In this chemical reaction, energy is absorbed which means it is an exothermic reaction so we can conclude that a soluble base is formed when sodium reacts with water.
Learn more about base here: https://brainly.com/question/3948796
The functional groups in an organic compound can frequently be deduced from its infrare d absorption spectrum. A compound, C3H6O2, exhibits intense absorption at 1740 cm-1, accompanied by a band at 1050 cm-1. No absorption above 3000 cm-1 is present .
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
The functional class(es) of this compound is(are) .(Enter letters from the table below, in any order, with no spaces or commas.)
a. alkane (List only if no other functional class applies.)
b. alkene
c. terminal alkyne
d. internal alkyne
e. arene
f. alcohol
g. ether
h. amine
i. aldehyde or ketone
j. carboxylic acid
k. ester
l. nitrile
Answer:
(b.) alkene
(i) aldehyde/ketone
(k.) ester
Explanation:
Peaks observed at:
1740 cm-1 indicates the presence of a carbonyl group: C=O aldehyde, ketone, esters.1050 cm-1 shows presence of carbo bonded to electronegative elements e.g. C-N or C-O3000cm-1 is usually indicative of alcohols or carboxylic acid -OH group, which rules out those classes.Question 2
Bromine will react very fast (almost instantly) with which compound?
O 1-Pentene
Cyclohexane
Heptane
Benzene
Answer:
1-Pentene
Explanation:
If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.
For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.
Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.
For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.
Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon dioxide was 93.7 grams, but the measured yield was 88.3 grams. What is the percent yield?
Answer:
Percent Yield = 94.237%
Explanation:
CO = Carbon Dioxide = Molar Mass 28g/mol
C = Carbon = 12g/mol
O = Oxygen = 16g/mol
Theoretical yield = 93.7 grams
Actual yield = 88.3 grams
Percent yield = (actual yield /theoretical yield )x100
Percent Yield = (88.3/93.7)x100
Percent Yield = 94.237%
100 mL of a buffer that consists of 0.20 M NH3 and 0.20 M NH4Cl is titrated with 25 mL of 0.20 M HCl. Calculate the pH of the resulting solution given that the Kb for NH3 is 1.8 x 10-5.
Answer:
pH = 9.03
Explanation:
The equilibrium of the NH₄Cl / NH₃ buffer in water is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Initial moles of both NH₃ and NH₄⁺ are:
0.100L ₓ (0.20 mol / L) = 0.0200 moles
The NH₃ reacts with HCl producing NH₄⁺, thus:
NH₃ + HCl → NH₄⁺ + Cl⁻
That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺
Moles added of HCl were:
0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:
NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles
NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.
Using H-H equation for bases:
pOH = pKb + log [NH₄⁺] / [NH₃]
Where pKb is -log Kb = 4.745.
Replacing:
pOH = 4.745 + log 0.0250mol / 0.0150mol
pOH = 4.967
As pH = 14- pOH
pH = 9.03
What is the molarity of a solution that has 4.4 moles BaCl2 in 1.5 Liters of solution?
Answer:
2.9 M
Explanation:
Step 1: Given data
Moles of barium chloride (solute): 4.4 moles
Volume of solution: 1.5 liters
Step 2: Calculate the molarity of barium chloride in the solution
The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.
[tex]M=\frac{4.4mol}{1.5L} =2.9 M[/tex]
Before running other simulations, try to think of solutions to this global energy dilemma. Write a paragraph in response to the following questions: Should we invest more heavily in renewables at a high up-front financial cost, and/or should we invest in finding more fossil fuels, and/or should we ignore the CO2 limit
Answer:
Yes, we should invest in renewable energies that decrease the levels of carbon dioxide in the atmosphere, that is, decrease the partial pressure of carbon dioxide in the environment since it is lethal for human life, in turn I consider that the exploitation of sources Oil was widely used but I think they should have an end, since they pollute oceans, oxygen, and fuels that are not necessary for today's technology since automobiles could be a base of electric or solar energy
Explanation:
Environmental pollution is a very serious current problem, since we should stop investing in oil sources and be able to keep those profits or that economy in another way or replace them with another resource, since if this greenhouse effect is not perceived as a current serious problem, we could run the risk of suffering natural catastrophes, systematic diseases, affections in the same life.
The total pressure of a mixture of H2, He, and Ar is 99.3 kPa. The partial pressure of the He is 42.7 kPa, and the partial pressure of Ar is 54.7 kPa. What is the partial pressure of hydrogen
Answer:
[tex]P_{H_2}=1.9kPa[/tex]
Explanation:
Hello,
Here, by using the Dalton's law, we can quantify the total pressure of a gaseous mixture by knowing the partial pressure of each gas, in case, hydrogen, helium and argon:
[tex]P_T=P_{H_2}+P_{He}+P_{Ar}[/tex]
In such a way, since we actually know the partial pressure of helium and argon, and the total pressure, we can compute the partial pressure of hydrogen as shown below:
[tex]P_{H_2}=P_T-P_{He}+P_{Ar}=99.3kPa-42.7kPa-54.7kPa\\\\P_{H_2}=1.9kPa[/tex]
Best regards.
At STP, what is the volume in milliliters of 0.0395 mol of fluorine gas, F2?
Answer:
884.8 mL
Explanation:
If we have STP condition (Standard temperature and pressure), we will have the following relationship:
[tex]1~mol=22.4~L[/tex] or [tex]1~mol=22400~mL[/tex]. With this in mind we can do the conversion:
[tex]0.0395~mol\frac{22400~mL}{1~mol}=884.8~mL[/tex]
In the 0.0395 mol of fluorine gas, we will have 884.8 mL of gas at STP conditions.
Reactions that undergo shifts in their equilibrium must be...
A.reversible reactions
B.chemical reactions
C.physical reactions
D.nuclear reactions
Answer:
A. Reversible reactions.
Explanation:
Reactions that undergo shifts in their equilibrium must be reversible reactions.
A reversible reaction is a chemical reaction that is capable of forming it's reactants back from the resulting formation of products. This simply means that, reversible reactions are chemical reactions that are in equilibrium because the forward and reverse path happens at the same rate.
For example, the reaction of hydrogen gas [tex]H_{2}[/tex] and iodine gas [tex]I_{2}[/tex] to form a chemical compound called hydrogen Iodide [tex]HI[/tex] is a reversible chemical reaction.
Forward reaction:
[tex]H_{2} + I_{2} ----> 2HI[/tex]
Reverse reaction:
[tex]2HI----> H_{2} + I_{2}[/tex]
A reversible reaction is denoted with a double arrow.
Also, reversible chemical reactions are controlled by the Le Chatelier's principle.
Is electricity an essential property of matter ?
Answer: I think so.. (sorry if i am wrong)
Explanation: Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.
An electrochemical cell contains a standard hydrogen electrode and a cathode consisting of a metallic chromium electrode, Cr(s), in contact with a 1.00 M chromium solution, Cr3 (aq). The voltage produced by this cell was measured at 25oC. Which statements describe the results of this measurement, assuming the conditions are ideal? The cell voltage with the appropriate sign equals
I. the cell potential.
II. the electromotive force.
III. the standard cell potential.
IV. the standard reduction potential for Cr3+/Cr.
a. I only
b. l and Il
c. I, II and III
d. I, II, II, and IV
e. ll only
Answer:
c. I, II and III
Explanation:
The cell is as follows
Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂
Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell
Cell potential = Ecathode - Eanode
= 0 - ( - E)
= E
E is cell potential and also standard cell potential or emf of the cell .
Standard reduction potential that is for Cr3+/Cr. is - E .
Hence statement I , II , III are right . IV th statement is wrong because of sign
Option c is correct.
The branch of chemistry which deals with electricity is called electrochemistry.
The correct answer is C
The cell representation is as follows:-
[tex]Cr / Cr^{2+}(1M) // H^+ ( 1 M ) / H_2[/tex]
The standard reduction potential of hydrogen cells is zero. Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell because it will behave as a cathode.
The formula of the electric cell is as follows :-[tex]Cell potential = E_{cathode} - E_{anode[/tex]
After putting the value the cell potential will be:-
Cell potential = 0 - ( - E)
The cell potential will be= E
The standard reduction potential that is for [tex]Cr^{3+}/Cr = - E .[/tex]
Hence, the correct option is C that is I, II, and IV.
For more information, refer to the link:-
https://brainly.com/question/25026730
Which of following changes that affect the composition of our atmosphere involve physical changes and which involve chemical reactions? Drag the appropriate items to their respective bins. View Available Hint(s) Reset Help Human activities such as the combustion of fossil fuels, has increased the levels of greenhouse At dew point, the water vapor begins to condense out of air. gases. Human activities, such as the combustion of fossil fuels, generate the aerosols. You observe hail when the temperatures are below the freezing point. Carbon dioxide is produced during respiration of plants, animals, fungi, and microorganisms. Chemical reaction Physical change
Answer:
1) Human activities such as the combustion of fossil fuels, has increased the levels of greenhouse gases---- Chemical reaction.
2) At dew point, the water vapor begins to condense out of air---- Physical changes.
3) Human activities, such as the combustion of fossil fuels, generate the aerosols---- Chemical reaction.
4) You observe hail when the temperatures are below the freezing point---- Physical changes.
5) Carbon dioxide is produced during respiration of plants, animals, fungi, and microorganisms----- Chemical changes.
Ideally, how should a buffer be prepared?
A. in such a way that pH = POH
B. in such a way that pH = pka
c. in such a way that pH = Kg
D. in such a way that pH = 1
Answer:
I may not be correct but i think its pH = Kg
Explanation:
Answer:
B. in such a way that pH = pKa
Explanation:
Ideally, the pH of the desired solution should have the same pKa as the pH, making the ratio 1:1.
Calculate the molar mass of a gaseous substance if 0.125 g of the gas occupies 93.3 mL at STP.
30.2 g/mol
30.4g/mol
30.6 g/mol
30.0 g/mol
None of the above
Answer:
30.0g/mol
Explanation:
Step 1: Given data
Mass of the gas: 0.125 gPressure (P): 1 atm (standard pressure)Temperature (T): 273.15 K (standard temperature)Volume (V): 93.3 mLStep 2: Calculate the moles of the gas
We will use the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1atm \times 0.0933L}{\frac{0.0821atm.L}{mol.K} \times 273.15K} = 4.16 \times 10^{-3} mol[/tex]
Step 3: Calculate the molar mass of the gas
4.16 × 10⁻³ moles correspond to a mass of 0.125 g. The molar mass of the gas is:
[tex]\frac{0.125g}{4.16 \times 10^{-3} mol} =30.0g/mol[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
Number of moles of Gas at STP,
[tex]\bold{n =\dfrac {PV}{RT}}[/tex]
where,
P - pressure
V- volume
R- gas constant
T - temperature
Put the values in the formula,
[tex]\bold{n =\dfrac {1 \times 0.0933} {0.082 \times 273.15 }}\\\\\bold{n =4.16 \timesw 10^-^3}[/tex]
The molar mass of the gas can be calculated using formula,
[tex]\bold {m = \dfrac {w}{n}}\\\\\bold {m = \dfrac {0.125} {4.16 \times 10^-^3}}\\\\\bold {m = 30g/mol}[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
To know more about molar mass, refer to the link:
https://brainly.com/question/12127540
Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.
Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
Where K of equilibrium is the Ka2: 6.5x10⁻⁸
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
pH = 7.1581516 mL of a 3.82 M sodium sulfate (Na2S04) solution is diluted with 0.875 L of water. What is the new concentration in molarity?
Answer
Molarity = [tex]1.4mol/L[/tex]
Explanation:
Molarity provides the number of moles of solute per liter of solution (moles/Liter). It is a means by which concentration of solution is measured.
SEE THE ATTACHMENT BELOW FOR STEP BY STEP SOLUTION.
Answer:
1.42 M
Explanation:
In this case have a dilution problem, therefore we need to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
What values we have?
[tex]C_1=3.82M[/tex]
[tex]V_1=526mL(0.516L)[/tex]
[tex]C_2=?[/tex]
[tex]V_2=?[/tex]
Now, we can calculate [tex]V_2[/tex] if we add the volumes, so:
[tex]0.516~L+~0.875~L=1.391~L[/tex]
So, [tex]V_2=1.391~L[/tex]
We can plug the values in the equation:
[tex]3.82~M*0.516~L=C_2*1.391~L[/tex]
[tex]C_2=\frac{3.82~M*0.516~L}{1.391~L}[/tex]
[tex]C_2=1.42~M[/tex]
I hope it helps!
A gas has a volume of 1140 ml at 37 ºC and 82.6 kPa pressure. Calculate its volume at STP.
Answer:
Volume, V2 at STP = 818.61ml.
Explanation:
Given the following parameters;
Volume, V1 = 1140ml
Volume, V2 =?
Temperature, T1 = 37ºC to Kelvin = 273+37 = 310K
Temperature, T2 = 273K is the standard temperature.
Pressure, P1 = 82.6kPa
Pressure, P2 = 101.3kPa is the standard pressure.
To solve for volume at standard temperature and pressure (STP), we would use the combined gas law;
The combined gas law is a combination of the other three gas laws, namely Gay Lusac's law, Boyle's law and Charles's law.
Combined gas law states that the ratio of the product or multiplication of volume and pressure to the temperature of a gas is equal to a constant.
Mathematically, [tex]PV/T = K[/tex]
P1V1/T1 = P2V2/T2
Let's make V2 to the subject formula;
Cross multiplying gives,
P1V1T2 = P2V2T1
Hence, [tex]V2 = (P1V1T2)/P2T1[/tex]
Substituting the parameters;
V2 = (82.6*1140*273)/101.3*310
V2 = 25706772/31403
V2 = 818.61ml.
An increase in the temperature of reactant causes an increase in the rate of reaction.
Which of the following is the best explanation?
a)the concentration of reactant increases
b) the activation energy decreases
c) the collision frequency increases
d) the fraction of collision with total kinetic energy larger than activation energy
increases
Answer:
d) The fraction of collision with total kinetic energy larger than activation energy increases.
Explanation:
Hello,
In this case, kinetic models explain how the rate of a chemical reaction is affected by several factors. In such a way, specifically for temperature, when it increases, the average velocity of the particles is also increased, for that reason, the collision frequency increases since the molecules are more likely to collide as they move faster and encounter to each other.
Nonetheless, it is the minor reason because the main reason is that the effective collisions increase when the temperature is increased, and they are related with the fraction of collision with total kinetic energy that turns out larger than the activation energy, therefore, answer is d).
Best regards.
Determine whether the following pairs of elements can form ionic compounds
Answer:
Oxygen and Magnesium.
Based on the difference in electronegativity, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, whereas lithium and calcium, sulfur and bromine, and manganese and chlorine cannot.
Ionic compounds are formed when there is a high electronegativity difference between two elements.
To determine whether the following pairs of elements can form ionic compounds, we need to look at their electronegativity difference:
1. Lithium and calcium: The electronegativity difference between Lithium and calcium is only 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
2. Sulfur and bromine: The electronegativity difference between sulfur and bromine is 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
3. Manganese and chlorine: The electronegativity difference between manganese and chlorine is 1.5, which is less than 1.7. Therefore, they cannot form an ionic compound.
4. Potassium and sulfur: The electronegativity difference between potassium and sulfur is 2.4, which is greater than 1.7. Therefore, they can form an ionic compound.
5. Chlorine and lithium: The electronegativity difference between chlorine and lithium is 2.8, which is greater than 1.7. Therefore, they can form an ionic compound.
6. Oxygen and magnesium: The electronegativity difference between oxygen and magnesium is 1.7, which is equal to 1.7. Therefore, they can form an ionic compound.
Therefore, based on the electronegativity difference between the elements, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, while lithium and calcium, sulfur and bromine, and manganese and chlorine cannot form ionic compounds.
Learn more about ionic compounds here:
https://brainly.com/question/9167977
#SPJ4
Which of the following isotopes of chlorine has 16 neutrons in its nucleus?
33Cl
38Cl
42Cl
Answer:
33Cl
Explanation:
The atomic mass of an element is made up of the proton and neutron.
chlorine has a constant number of 17 protons.
33-17 = 16
1. the purpose of the aqueous solutions in a galvanic cell is to?
a. provide ions to be oxidized and reduced
b. provide a path for the free flow of electrons
c. maintain charge balance in the cell
d. correct any volume changes in the cell
2. given these half-reactions,
B2 + 2e- -> 2B- Ecell= 0.662 V
A+ Ie- -> A Ecell= -1.305V
what is the standard potential for the overall reaction. 2A+B2 -> 2AB
a. -1.97 V
b. -0.643 V
c. +3.272 V
d. +1.967 V
Answer:
1. The correct option is;
c. maintains charge balance in the cell
2. The correct option is;
c. +3.272 V
Explanation:
The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral
Therefore, the aqueous solution maintains the charge balance in the cell
2. Here we have;
B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V
A⁺ + 1e⁻ → A Ecell = -1.305 V
Hence for the overall reaction, we have;
2A + B₂ → 2AB gives;
(0.662) - 2×(-1.305) = +3.272 V.
What are the differences between cholesterol and ergosterole?
An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Answer:
101
Explanation:
Provided that
[tex]S_1 = S_2 = same\ V_{max}[/tex]
And,
[tex]S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM[/tex]
Now we expect the same
{S} (0.1mM)
This determines that [tex]S_1[/tex] generates a higher rate of product formation as compared to the [tex]S_2[/tex]
So we can easily calculate the [tex]V_{max}[/tex] for either of [tex]S_1[/tex] or [tex]S_2[/tex] as we know that Tube 1 is [tex]S_2[/tex] and tube 2 is [tex]S_1[/tex]
As we know that
[tex]V_0 = V_{max}\ {S} / (K_M + {S})[/tex]
As the rates do not include any kind of units so we do not consider the units for [tex]V_{max}[/tex]
Now the calculation is
[tex]0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)[/tex]
[tex]V_{max} = 0.5 (20.1\ mM) / 0.1\ mM[/tex]
= 100.5
≈ 101
How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Answer:
Grams of mercury= 0.06 g of Hg
Note: The question is incomplete. The complete question is as follows:
A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Explanation:
Since one fluorescent light bulb contains 4 mg of mercury,
15 such bulbs will contain 15 * 4 mg of mercury = 60 mg
1 mg = 0.001 g
Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.
Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.
Total amount of mercury in 15 compact fluorescent light bulbs is 0.06 gram of mercury.
Compact fluorescent light bulbs and mercury:What information do we have?
Number of compact fluorescent light bulbs = 15 bulb
Amount of mercury in each bulb = 4 mg
Total amount of mercury = Number of compact fluorescent light bulbs × Amount of mercury in each bulb
Total amount of mercury = 15 × 4
Total amount of mercury = 60 mg
Total amount of mercury = 60 / 1000
Total amount of mercury = 0.06 gram of mercury
Find out more information about 'Mercury'.
https://brainly.com/question/2279846?referrer=searchResults
Select the atoms or ions drawn with valid Lewis dot structures. A) A carbon has a dot on top, right, bottom and to the left.a nitrogen has one dot on top, left and to the bottom and has a charge of minus 3.a nitrogen has a dot on top, right, bottom and to the left. B) An oxygen has two dots on top and bottom and one dot to the left and to the right. C) A carbon has two dots on top, right, bottom and to the left and a charge of plus four. D) An oxygen has two dots on top, left and to the bottom and a charge of minus 2.
Answer:
B, C
Explanation:
The atoms or ions with the valid Lewis dot structures are B and C.
In A;
The Lewis structure of the carbon is correct. Each of the four dots represent the four valence electrons.
The nitrogen with one dot on top, left and to the bottom and has a charge of minus 3 is wrong. For it to have a charge of -3 it must have 8 lewis dots ( two on the top, right, bottom and to the left)
The nitrogen with four dots (on top, right, bottom and to the left) is wrong.
In B;
An oxygen has two dots on top and bottom and one dot to the left and to the right. This is correct , the 6 dots represent the valence electrons of oxygen.
In C;
A carbon has two dots on top, right, bottom and to the left and a charge of plus four. This is correct because the charge indicates that it has gained four extra electrons so its valence electrons is now 8.
In D;
An oxygen has two dots on top, left and to the bottom and a charge of minus 2. This is wrong because the lewis dots are incomplete. Two dots are missing.
Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ Express your answer numerically, to four significant figures and in terms of kJ.
Answer:
ΔH = 130.5 kJ
Explanation:
Hello,
In this case, by using the Hess law, we compute the enthalpy of the required reaction:
C(s) + H2O(g) --> CO(g) + H2(g)
Thus, the first step is to keep the following reaction unchanged:
C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ
Next, we invert and halve this reaction:
2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ
So the enthalpy of reaction is inverted and halved:
CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ
Then, we also invert and halve this reaction:
2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ
So the enthalpy of reaction is inverted and halved as well:
H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ
Finally, we add the three reactions to obtain the required reaction:
= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)
= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)
= C (s) + H2O → H2 (g) CO (g)
So enthalpy is computed by:
ΔH = -393.5 kJ + 283 kJ + 241.8 kJ
ΔH = 130.5 kJ
Best regards.
Considering the Hess's Law, the enthalpy change for the reaction is 131.3 kJ.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
C(s) + H₂O(g) → CO(g) + H₂(g)
You know the following reactions, with their corresponding enthalpies:
Equation 1: C (s) + O₂(g) → CO₂ (g) ΔH = -393.5 kJ
Equation 2: 2 CO (g) + O₂ (g) → 2 CO₂ (g) ΔH= -566.0 kJ
Equation 3: 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ΔH= -483.6 kJ
First stepFirst, to obtain the enthalpy of the desired chemical reaction you need one mole of C(s) on reactant side and it is present in first equation so let's write this as such.
Second stepNow, 1 mole of CO(g) must be a product and is present in the second equation. Since this equation has 2 moles of CO(g) on the reactant side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of CO(g).
When an equation is inverted, the sign of ΔH also changes.
And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also is divided by 2.
Third step
Finally, 1 mole of H₂O(g) must be a reactant and is present in the third equation. Since this equation has 2 moles of CO(g) on the product side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of H₂O(g).
So, the sign of ΔH also changes and the variation of enthalpy is divided by 2.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: C (s) + O₂(g) → CO₂ (g) ΔH = -393.5 kJ
Equation 2: CO₂ (g) → CO (g) + [tex]\frac{1}{2}[/tex] O₂ (g) ΔH= 283 kJ
Equation 3: H₂O (g) → H₂ (g) + [tex]\frac{1}{2}[/tex] O₂ (g) ΔH= 241.8 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
C(s) + H₂O(g) → CO(g) + H₂(g) ΔH= 131.3 kJ
Finally, the enthalpy change for the reaction is 131.3 kJ.
Learn more:
brainly.com/question/5976752?referrer=searchResults brainly.com/question/13707449?referrer=searchResults brainly.com/question/13707449?referrer=searchResults brainly.com/question/6263007?referrer=searchResults brainly.com/question/14641878?referrer=searchResults brainly.com/question/2912965?referrer=searchResults43. Calculate the equilibrium constant at the temperature given. (a) O2 (g) + 2F2 (g) ⟶ 2F2 O(g) (T = 100 °C) (b) I2 (s) + Br2 (l) ⟶ 2IBr(g) (T = 0.0 °C) (c) 2LiOH(s) + CO2 (g) ⟶ Li2CO3 (s) + H2 O(g) (T = 575 °C) (d) N2 O3 (g) ⟶ NO(g) + NO2 (g) (T = −10.0 °C) (e) SnCl4 (l) ⟶ SnCl4 (g) (T = 200 °C)
Answer:
Explanation:
O₂(g) + 2F₂(g) ↔ 2F₂(g)
Stabdard ΔG values are
[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]
[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]
[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]
[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]
Now,
[tex]\Delta G^\circ =-RTInK[/tex]
Given T = 100°C
= 100+ 273.15 = 373.15K
R = 8.314J/k / mol
so,
83800 = -8.314 * 373.15 * InK
InK = -27.0116
K = 1.858 * 10⁻¹²
Equilibrium constant = 1.858 * 10⁻¹²When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to effuse as the same amount of CO2 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus chloride
Answer:
molar mass of the phosphorus chloride = 138.06 g/mol
Explanation:
mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.
mass = 3.51 g
lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec
From this we can say that
rate of effusion of CO2 = 3.51/1 = 3.51 g/s
rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s
From graham's equation of effusion,
[tex]\frac{Rc}{Rp}[/tex] = [tex]\sqrt{\frac{Mp\\}{Mc} }[/tex]
Rc = rate of effusion of CO2 = 3.51 g/s
Rp = rate of effusion of phosphorus chloride = 1.98 g/s
Mc = molar mass of CO2 = 44.01 g/mol
Mp = molar mass of the phosphorus chloride = ?
Imputing values into the equation, we have
[tex]\frac{3.51}{1.98}[/tex] = [tex]\sqrt{\frac{Mp\\}{44.01} }[/tex]
1.77 = [tex]\frac{\sqrt{Mp} }{6.64}[/tex]
11.75 = [tex]\sqrt{Mp}[/tex]
Mp = [tex]11.75^{2}[/tex]
Mp = molar mass of the phosphorus chloride = 138.06 g/mol