Answer:
10
Explanation:
The 5 will multiply the subscript 2 on the K atoms making it 10
describe how the relative density of a salt solution can be determined using the hydrometer
The hydrometer is an instrument used to determine the density of liquids, and looks like a large glass float filled at the bottom with grains of lead or mercury.
Because the mass of the hydrometer is known and thus the mass of the displaced liquid, the density can be determined using the volume.
There is a scale on the handle, the density can be read here. The deeper the hydrometer sinks, the lower its density.
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Burning a compound of calcium, carbon and nitrogen in oxygen generates calcium oxide (CaO), carbon dioxide (CO2) and nitrogen dioxide (NO2). A small sample gives 3.106 g CaO, 2.439 g CO2, and 5.097g NO2. Determine the empirical formula of the compound.
Answer:
CaCN₂
Explanation:
A compound of calcium, carbon and nitrogen CaₐCₓNₙ in oxygen will burning producing:
CaₐCₓNₙ + O₂ → aCaO + xCO₂ + nNO₂
Moles of the oxides CaO, CO₂, NO₂ are:
CaO: 3.106g ₓ (1 mole / 56.08g) = 0.0554 moles of CaO = moles of Ca
CO₂: 2.439g ₓ (1mole / 44g) = 0.0554 moles CO₂ = moles C
NO₂: 5.097g ₓ (1mole / 46g) = 0.111 moles NO₂ = moles N
Empirical formula is the chemical formula that represents the simplest ratio of elements in a compound.
Having as basis 0.0554 moles (The lower number of moles):
0.0554 moles Ca / 0.0554 = 1 Ca
0.0554 moles C / 0.0554 = 1 C
0.111 moles N / 0.0554 = 2N
Thus, the compound CaₐCₓNₙ has as empirical formula:
Ca₁C₁N₂ = CaCN₂
Imagine you are volunteering in a 1st grade classroom. The students recently learned about the water cycle. One student raises her hand and asks, "I heard that elements like carbon can cycle too. How does carbon cycle? Is it like how water cycles?" How would you explain the cycling of carbon to this 1st grade class? Keep in mind that the students have likely not taken an advanced science course yet!
Answer:
Carbon cycle is also important for the survival of life on earth.
Explanation:
In carbon cycle, carbondioxide gas is present in the atmosphere. This carbondioxide gas is taken inside by the plant body through stomata and made food from it. When this food is eaten by animals, this carbon is transferred to their bodies. During breathing, carbondioxide gas is removed from animal body and the carbondioxide gas again goes to the atmosphere.
3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the concentration of SO2Cl2 were monitored as shown in the table.
SO2Cl2 (g) SO2 (g) + Cl2 (g)
Time (min) Conc. of SO2Cl2 (mol/L)
0 0.1000
100 0.0876
200 0.0768
300 0.0673
400 0.0590
500 0.0517
600 0.0453
700 0.0397
800 0.0348
900 0.0305
1000 0.0267
1100 0.0234
a) Determine graphically whether the kinetics of the reaction is zero order, first order or second order with respect to SO2Cl2 and then write the rate equation.
b) Determine the rate constant (k) of the reaction.
c) Determine the half-life (t½) for the reaction.
d) What will be the concentration of SO2Cl2 left in the reaction mixture at 1500 minutes?
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:
[tex]m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}[/tex]
c) Half life for first-order kinetics is computed by:
[tex]t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min[/tex]
d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:
[tex]C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M[/tex]
Best regards.
The bitter taste often associated with leafy green vegetables such as spinach, parsley, and cilantro is often due to oxalic acid, which has a molecular weight of 90.04g. The percentage of the acid present can be determined through an acid-base titration with NaOH once the sample has been prepared and the oxalic acid is extracted into an aqueous solution. A 100.0-g sample was treated so that all of the oxalic acid dissolved. Because the solution was intensely green due to chlorophyll, an indicator could not be used, and the titration was monitored using pH meter. The following data were obtained. Concentration of NaOH -0.693 M volume 0.00 pH 1.63 2.02 14.00 1.01 12.00 2.49 2.46 10.00 4.56 2.80 8.00 6.01 2.98 pH 7.69 3.16 6.00 8.33 3.23 4.00 3.51 10.72 13.35 2.00 3.91 14.14 4.09 0.00 0 5 10 1516 20 15.41 4.67 15.67 4.97 Volume of NaOH 15.85 5.49 5.79 15.89 (a) How many moles of oxalic acid were present in the 100- g sample? 15.91 6.09 15.95 12.14 16.01 12.74 16.52 13.61 17.06 13.65 (b) Calculate the percentage of oxalic acid in the spinach.
Answer:
1. Moles of oxalic acid in 100 g sample = 0.0055moles
2. Percentage oxalic acid in 100g sample = 0.5%
Explanation:
Check attachment below for explanation and correct chart arrangement
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of water is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of water. Round your answer to significant figures.
Answer:
See explanation below
Explanation:
You are missing certain data. I found one, in the attached picture. All you have to do is replace your data in this procedure to get an accurate answer.
According to the picture attached, we have 2.4 g of HBr and 1.9 g of NaOH. This mix produces 0.411 g of water. To get the %yield of water, we first need to see how many theorical grams of water are formed. With this we can calculate the %yield.
The overall reaction is:
HBr + NaOH --------> NaBr + H₂O
We have a mole ratio of 1:1 so, let's see which is the limiting reactant between the acid and the base, using the reported molar mass for each (MM HBr = 80.9119 g/mol; MM NaOH = 39.997 g/mol)
moles HBr = 2.4 / 80.9119 = 0.02966 moles
moles NaOH = 1.9 / 39.997 = 0.0475 moles
So, we can clearly see that the moles of NaOH are in excess, so the HBr is the limiting reactant.
As we have a mole ratio of 1:1 with each compound here, and that the HBr is the limiting reactant we can say that:
moles HBr = moles H₂O = 0.02966 moles of water
Now, using the molar mass of water, let's calculate the theorical yield of water: (MM water = 18.01528 g/mol
m H₂O = 0.02966 * 18.01528 = 0.534 g
We have the theorical yield, let's calculate the % yield:
% = 0.411 / 0.534 * 100
% = 76.92%
Magine that you have a metal bar sitting half in the Sun and half in the dark. On a sunny day, the part of the metal that has been sitting in the Sun feels hot. If you touch the part of the metal bar that has been sitting in the dark, will it feel hot or cold?
Answer:
The metal will feel hot.
Explanation:
Metals are known for their high thermal conductivity. If one part of the metal feels hot, the other will as well.
_______________________________________________________
This piece of metal has been exposed to the sun, and will feel hot after a certain amount of time. By the statement above, the unexposed part will feel hot as well.
Hope that helps!
Please show work and explain:
The reaction A + B → C is studied similarly to our study of phenolphthalein fading kinetics.
The corresponding rate law is Rate = k[A]m[B]m
In this particular experiment, the concentration of A is sufficiently high that the pseudo-order rate law: Rate = k1[B]n can be written, where k1 = k[A]m.
If the study establishes that the reaction is first order in B (n = 1) and that the pseudo rate constant k1 has the following values at varying concentrations of B, what is the order of the reaction in reactant A (what is the value of m)?
k1 [B]
0.0034 0.200 M
0.0020 0.100 M
0.0010 0.050 M
Options:
A. 0
B. 1
C. 2
D. 3
Answer:
Explanation:
Consider the given reaction,
A+B +C
[tex]rate=k[A]^m[B]^m[/tex]
Reaction is first order with respect to reactant B.
The pseudo order rate is
[tex]rate=k_1 [B]^m,\\\\where k_1 =k[A]^m[/tex]
Pseudo order reaction is the reaction in which the reactant concentration remains constant; its concentration is very high so it does not involve in the rate of the reaction. Here, the concentration of A is sufficiently high so this reaction becomes the pseudo first order reaction.
The reaction is first order with respect to [B] with rate constant k1 .
Using the given table, concentration of B is decreasing the rate also decreasing
Therefore, order of the reaction with respect to A is 1.
Value of m is 1.The order of the reaction in reactant A is 1, means the value of m for given rate should be 1.
What is pseudo order reaction?Pseudo order reactions are those reactions in which two reactants are present but the concentration of one of the reactant is present in excess quantity and it seems like first order reaction.
Given chemical reaction is: A + B +C
According to the question, given reaction is first order with respect to the reactant B and rate law will be represented as:
Rate = k[A]m[B]m
Also given that reactant A is present in high concentration in the reaction, so the rate of pseudo order reaction is:
Rate = k1[B]n, where
k1 = k[A]m
According to the given table with respect to concentration of reactant B and rate k1, it is clear that rate of the reaction decreases as the concentration of reactant B decreses. So, to get the required result order of the reaction or value of m with respect to the reactant A should be 1.
Hence option (B) is correct i.e. 1.
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3. The exosphere is the outer layer of the
O D. thermosphere
OC. mesosphere
O A. ozone layer
O B. troposhpere
ang
AT
HHH
Answer:
D. exosphere is the outer layer of the thermosphere
A discharging lead-acid battery is best described as
1.) chemical cells that use an electrical current
2.) chemical cells that produce an electrical current
3.)electrolytic cells that use an electric current
4.)electrolytic cells that produce an electric current
Answer:
chemical cells that produce an electrical current
Explanation:
lead acid battery is a type of chemical cell, that discharges to produce electrical current.
lead acid battery uses lead peroxide for the conversion of the chemical energy into electrical power when it is discharging. It is also rechargeable, that is, it can convert electrical energy back to chemical energy during charging.
Thus, a discharging lead acid battery is best described as a chemical cells that produce an electrical current
Answer:
The correct option is;
2.) Chemical cells that produce an electric current
Explanation:
A discharging lead-acid battery is an electrochemical cell that produces electrical energy through the chemical reaction taking place at the battery electrodes which have lead as the anode and lead oxide as the cathode which result in the flow of an electric current, thereby converting chemical energy of the lead, lead oxide and aqueous sulfuric acid system into electrical energy, therefore, it is a chemical cell that produces an electric current.
what is the mass of 3.50 moles of silicon dioxide
Answer:
210. g
Explanation:
SiO2 molar mass=60.09 g/mol
3.50 mol * 60.09 g/mol = 210. g
Many computer chips are manufactured from silicon , which occurs in nature as SiO2. WhenSiO 2 is heated to melting , it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of SiO2 reacts with 78.3 kg of carbon to produce 66.1 kg of silicon. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.
Answer:
In the given case, SiO2 is the limiting reactant, 71.72 Kg is the theoretical yield, and 90.8 percent is the experimental yield.
Explanation:
A reactant that completely gets used up in a chemical reaction, and it limits the concentration of product, which can be produced is termed as a limiting reactant. The maximum concentration of product that can get generated from the limiting reactant is termed as the theoretical yield.
Based on the given question, the balanced chemical reaction is:
SiO₂ + 2C ⇒ Si + 2CO
Now the mole ratio of SiO₂: C = 1: 2
Based on the given information, the mass of SiO₂ is 155.8 Kg. The no. of moles can be calculated by using the formula, n = weight/molecular mass.
The molecular mass of SiO₂ is 60.08 gram per mole. By putting the values we get,
n = 155.8 × 10³ grams / 60.08 grams/mol
n = 2.59 × 10³ moles
The weight of Carbon given is 78.3 Kg or 78.3 × 10³ grams, the molecular mass of C is 12 gram per mole. Similarly, the moles of C will be,
n = 78.3 × 10³ grams / 12 g/mol
n = 6.52 × 10³ moles
The experimental molar ratio of SiO₂: C is,
= 2.59 × 10³: 6.52 × 10³
= 2.59: 6.52 = 1: 2.5
Hence, it is clear that carbon in the given case is present in excess amount, therefore, SiO₂ is a limiting reactant.
With the help of the balanced equation, the molar ratio of SiO₂: Si is 1: 1. As SiO₂ is the limiting reactant, therefore, theoretical yield of the reaction will be determined by SiO₂.
As calculated, the moles of SiO₂ used is 2.59 × 10³
Hence, the moles of Si produced in the given reaction is 2.59 × 10³.
Now the mass of Si produced theoretically can be determined by using the formula, moles = weight/molar mass. The molar mass of Si is 28.08 gram per mole. Now putting the values we get,
Weight = 2.59 × 10³ moles × 28.08 gram per mole
Weight = 72.72 × 10³ grams or 72.72 Kg
The theoretical yield of Si is 72.72 Kg, however, the experimental yield is 66.1 Kg. Therefore, the percent yield for the reaction will be,
= Experimental yield / Theoretical yield × 100 %
= 66.1 Kg / 72.72 Kg × 100%
= 90.8 %
Methanol CH3OH is the simplest of the alcohols . It is sinthesized by the reaction of hydrogen and carbon monoxide.
CO + H2 -----------> CH3OH
a) If 500 ml CO and 750 mol H2 are present, which is the limiting reactant?
b) How many mols of excess reactant remain unchange?
c) How many moles of CH3OH are formed ?
Answer:
a) CO is the limiting reactant.
b) [tex]n_{H_2}^{excess}=749.9592molH_2[/tex]
c) [tex]n_{CH_3OH}=0.0204molCH_3OH[/tex]
Explanation:
Hello,
In this case, the balanced chemical reaction:
[tex]CO + 2H_2 \rightarrow CH_3OH[/tex]
So we proceed as follows:
a) We first compute the moles of CO by using its density (0.00114 g/mL) and molar mass (28 g/mol):
[tex]n_{CO}=500mLCO*\frac{0.00114gCO}{1mLCO}*\frac{1molCO}{28gCO}=0.0204molCO[/tex]
Next, with the given 750 moles of hydrogen, we can compute the moles of carbon monoxide that are consumed by such amount of hydrogen by using their 1:2 molar ratio:
[tex]n_{CO}^{consumed\ by\ H_2}=750molH_2*\frac{1molCO}{2molH_2}=375molCO[/tex]
Thus, we see a clear excess of hydrogen, for that reason the carbon monoxide is the limiting reactant.
b) In this case, we first compute the moles of CO that are not consumed:
[tex]n_{CO}^{unchanged}=375mol-0.0204mol=374.9796molCO[/tex]
Next, we use the 1:2 molar ratio again to compute the unchanged moles of hydrogen which is the excess reactant:
[tex]n_{H_2}^{excess}=375.9796molCO*\frac{2molH_2}{1molCO} =749.9592molH_2[/tex]
c) Finally, we use the reacting moles of carbon monoxide to compute the formed moles of methanol by using the 1:1 molar ratio:
[tex]n_{CH_3OH}=0.0204molCO*\frac{1molCH_3OH}{1molCO} =0.0204molCH_3OH[/tex]
Best regards.
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder that contains 12.13 mL H2O. What will be the final volume reading on the graduated cylinder?
Answer:
12.54mL
Explanation:
Step 1:
Data obtained from the question:
Mass of Zn = 2.95g
Density of Zn = 7.14 g/mL
Volume of water = 12.13mL
Final volume reading of cylinder =..?
Step 2:
Determination of the volume of Zn. The volume of Zn can be obtained as follow:
Density = Mass /volume
7.14 = 2.95/volume
Cross multiply
7.14 x Volume = 2.95
Divide both side by 7.14
Volume = 2.95/7.14
Volume of Zn = 0.41mL
Step 3:
Determination of the final volume reading of the cylinder.
The final reading of the cylinder will be the addition of the volume of water and the volume of Zn as shown below:
Final volume of cylinder = volume of water + volume of Zn.
Volume of water = 12.13mL
Volume of Zn = 0.41mL
Final volume =..?
Final volume of cylinder = 12.13 + 0.41
Final volume of cylinder = 12.54mL
Therefore, the final volume reading of the cylinder is 12.54mL
The final volume reading of the graduated cylinder is 12.54 mL.
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder. The volume of the piece of zinc is:
[tex]2.95 g \times \frac{1mL}{7.14 g} = 0.413 mL[/tex]
The final volume reading of the graduated cylinder will be equal to the sum of the volumes of the piece of zinc and the water.
[tex]V = V(H_2O)+V(Zn) = 12.13 mL + 0.413 mL = 12.54 mL[/tex]
The final volume reading of the graduated cylinder is 12.54 mL.
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If 0.089 grams of KI is dissolved in 500g of H2O, what is the concentration of the resulting solution in parts per million?
Answer:
The concentration of the resulting solution in parts per million is 177.97
Explanation:
Parts per million (ppm), is a unit of measure for concentration that refers to the number of units of the substance per million units of the set.
The concentration in parts per million expressed in mass / mass is calculated by dividing the mass of the solute (ms) by the mass of the solution (md, sum of the mass of the solute and the mass of the solvent), both expressed in the same unit and multiplied by 10⁶ (1 million).
[tex]ppm=\frac{ms}{md} *10^{6}[/tex]
So, being:
ms: 0.089 grams of KImd: 0.089 grams of KI + 500 grams of H₂O= 500.089 gramsReplacing:
[tex]ppm=\frac{0.089 grams}{500.089 grams}*10^{6}[/tex]
ppm= 177.97
The concentration of the resulting solution in parts per million is 177.97
One kilogram is the same as
Answer:
1000 grams
Explanation:
Kilo- is the SI prefix meaning 1000. So, 1 kilogram is the same as 1000 grams.
if two atoms are bonded to a central atom with no lone pairs , how will they be arranged?
Methanol is classified as what alcohol
Answer:
Hey mate,
Here is your answer. Hope it helps you
Explanation:
Primary alcohols are those alcohols where the carbon atom of the hydroxyl group(OH) is attached to only one single alkyl group. Some of the examples of these primary alcohols include Methanol (, propanol, ethanol, etc. The complexity of this alkyl chain is unrelated to the classification of any alcohol considered as primary. The existence of only one linkage among –OH group and an alkyl group and the thing that qualifies any alcohol as a primary.
A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128 J/g oF; water = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium
Answer: The final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
Explanation:
The given data is as follows.
mass = 7.62 g, [tex]T_{2} = 10.8^{o}C[/tex]
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = [tex]mC \times \Delta T[/tex]
= [tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex]
Now, heat gained by lead will be calculated as follows.
q = [tex]mC \times \text{Temperature change of lead}[/tex]
= [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
According to the given situation,
Heat lost = Heat gained
[tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex] = [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
T = [tex]50.26^{o}C[/tex]
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
which statement best describle for formation of solution
Explanation:
small amount of solute is dissolved in a large amount of solvent. it would not make sense to have more solute and solvent. you can think of it like this, you can dissolve a teaspoon of salt in a litre water but you can't resolve 1 kg of salt in a teaspoon of water.
also, the solute is what is begin dissolved and the solvent is what dissolving the solute so the one day the solvent is dissolved in the solute does not make complete sense.
QUESTION 5
The attraction between the nucleus of the atom and the valence electrons for any given atom is called what?
Answer:
Electrostatic Force of Attraction
Hope this helps ;)
Explanation:
A primary amine contains a five-membered carbon ring. The degree of unsaturation for the five-membered carbon ring is equal to three. In the chemical formula, there are seven carbon atoms. Determine its chemical formula.
Answer:
C₇H₁₁N
Explanation:
To find the chemical formula, first, let's evaluate the given conditions.
1. Is a primary amine:
This means that the amine is connected to a carbon, like NH₂-R
2. The degree of unsaturation of the five-membered carbon ring is equal to three:
One of the degrees of unsaturation is for the five-membered carbon ring, the other two are for pi-bonds. This means that the five-membered carbon ring has 2 pi-bonds inside.
3. There are seven carbon atoms:
Five of these seven carbons are for the carbon ring and the other two are bonded to the primary amine.
From all of the above, we have that the chemical formula is:
C₅H₅-C₂H₄-NH₂ = C₇H₁₁N
I hope it helps you!
How sensitive to changes in water temperature are coral reefs? To find out, measure the growth of corals in aquariums where the water temperature is controlled at different levels. Growth is measured by weighing the coral before and after the experiment.
(a) What are the explanatory and response variables?
explanatory variable:
response variable:
(b) Are they quantitative or categorical?
The explanatory variable is:
The response variable is:
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.
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At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25°C and must be heated to 100°C to make the balloons float. 1.00kg of butane C4H10 fuel are available to be burned to heat the air. Calculate the maximum number of balloons that can be inflated with hot air.
Answer:
The correct answer is 1.5 balloons.
Explanation:
Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,
= 4/3 × π × (170)³ cm³
= 20.5176 × 10⁶ cm³
= 20.5176 m³
The density of air at 100 degree C s 0.946 Kg m⁻³
The mass of air in each balloon can be calculated by using the formula,
Mass = density × volume
Mass = 0.946 Kg m⁻³ × 20.5176 m³
Mass = 19.410 Kg
The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,
= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C
= 14.68 × 10⁵ J
The concentration of butane given is 1.00 Kg or 1000 g
The molecular weight of butane is 58.12 g per mole
The moles or n can be calculated by using the formula,
n = mass / mol.wt
n = 1000 g/58.12 g/mol = 17.20 mol
The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,
= 125.7 × 10³ J/mol × 17.20 mol
= 2.16 × 10⁶ J
The number of balloons that can be inflated with hot air is,
= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon
= 1.5 balloons
Hence, maximum of 1.5 balloons can be inflated.
Why do heart diseas patient's eat oil instead of fat?
Answer:
lube
Explanation:
Sodium Chloride (NaCl) has a molecular mass of 58 grams/mole. What is the molarity of a NaCl solution with a
concentration of 232 grams/liter?
2.0 molar
4.0 molar
0.5 molar
3.0 molar
Answer:
2nd option
Explanation:
Molarity is the number of moles of the solute (NaCl) in 1 litre of the solution (NaCl solution).
Given: concentration= 232g/ L
what we are trying to achieve is __mol/ L.
So in 1 litre, we have 232g of NaCl.
To convert mass to mole, we divide it by the Mr.
Given that the Mr is 58g/mol,
number of moles
= 232 ÷58
= 4
Thus, 1 litre has 4 moles of NaCl.
Therefore, the molarity is 4.0 mol/L.
Eduardo hurts his knee during a basketball game. The trainer applies a cold pack, which gets cold after being squeezed. The next day, Eduardo’s friend Beth is going sledding. Beth buys some hand warmers and puts them into her gloves. All morning, her hands stay toasty warm.
Required:
a. How do you think these devices work?
b. Where do you think the "cold" and the heat comes from?
Answer:
How do you think these devices work?
Both devices follows the principle and endothermic and exothermic reactions.The cold pack is made up of two bags. The first bag contains water.And it is placed inside the second bag which contains NH4NO3 , and urea. Therefore, when the water bag is squeezed, the bag busted releasing the water which dissolves the surrounding solid compounds. These compounds absolved heat energy in an endothermic reaction.
The components of hand warmers are activated carbon, salt iron, water,,and cellulose.The Oxidation of iron when exposed to the air (oxygen)generated heat in an exothermic reaction to keep the hand warm.
Thus the cold results from the endothermic reaction as the solid compounds in cold packs absorbed water to break bonds to melts,and the hotness comes from formation of new bonds(exothermic) as oxidation of Fe to form new compounds with other chemical occurs.
Explanation:
What is the molarity of a solution of 3 moles (mol) of FeBr3 in 1/2 L water?
Answer:
3mol/0.5liters or 6 M
Explanation:
The molarity of a solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
What is Molarity ?Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the number of moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution.
Molarity is number of moles of solute in 1 litre of solution.
Therefore,
If 0.5 ltr of solution consisting of 3 moles of FeBr₃
Thus,
1 ltr of solution consisting of 6 moles of FeBr₃.
6 moles of solute in 1 ltr of solution means 6 Molar solution.
Hence, The molarity of a given solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
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PLEASE HELP I HAVE LIMITED TIME!!
How many particles of water (H20) are in a collection of snowflakes
with a mass of 0.005 g?
Answer:
Explanation:
Si tomamos en cuenta el peso molecular del agua, que es equivalente a:
1 Átomo de H₂O
O = 16 gr/mol
H = 1 gr/mol
H₂O = 18 gr/mol
Teóricamente sabemos que en 1 mol de H2O habrá 18 gr.
Para obtener los moles presentes en 1 mg de H₂O, (como 1 gr = 1000 mg), decimos:
1 mol H2O ………………………….. 18000 mg
X …………………………… 1 mg
X = 1 / 18000 = 5,56 X 10⁻⁵ moles de H20
Y para obtener la cantidad de moléculas presentes, de acuerdo a los moles, multiplicamos por el número de Avogadro (6,023 X 10²³ moléculas /mol)
Moléculas de H₂O = 5,56x 10⁻⁵ mol x 6,023 x 10²³
Moléculas de H₂O = 3,34488 x10¹⁹ moléculas de H₂O
En el copo de nieve habrá 3,34488 x 10¹⁹ moléculas de H₂O.
Espero que te sirva =)
5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
What do you mean by mole ?The term mole is defined as the amount of substance of a system which contains as many elementary entities.
One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.
The mole concept can be used to convert between mass and number of particles.
1 atom of H₂O
O = 16 g/mol
H = 1 gr/mol
H₂O = 18 gr/mol
1 mol H₂O ………………………….. 18000 mg
? X …………………………… 1 mg
X = 1 / 18000
= 5,43 X 10²² moles de H20
Thus, 5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
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Select the precipitate that forms when aqueous ammonium sulfide reacts with aqueous copper(ii) nitrate. refer to the solubility rules. Provide the complete ionic equation, net ionic equation and spectator ions for https://tex.z-dn.net/?f=%28NH_4%29_2S%28aq.%29%2BCu%28NO_3%29_2%28aq.%29%5Crightarrow%20CuS%28s%29%2B2NH_4NO_3%28aq.%29
Answer:
- The precipitate that forms from the reaction of Ammonium Sulfide and Copper (II) Nitrate is the insoluble Copper (II) Sulfide (CuS).
- Overall ionic equation
2NH₄⁺ + S²⁻ + Cu²⁺ + 2NO₃⁻ → 2NH₄⁺ + 2NO₃⁻ + CuS (s)
- Net Ionic equation
S²⁻ + Cu²⁺ → CuS (s)
Or better written as
Cu²⁺ + S²⁻ → CuS (s)
Spectator ions include 2NH₄⁺ and 2NO₃⁻.
Explanation:
Ammonium Sulfide - (NH₄)₂S
Copper (II) Nitrate - Cu(NO₃)₂
The overall equation for this reaction is a double displacement reaction and the balance form is represented by
(NH₄)₂S (aq) + Cu(NO₃)₂ (aq) → 2NH₄NO₃ (aq) + CuS (s)
Copper (II) Sulfide (CuS) is the insoluble precipitate formed from this reaction.
Note that the compounds in aqueous form are the ones that can allow the existence in ionic form, the insoluble solid compound stays in that form without ionizing, hence, the overall ionic equation is
2NH₄⁺ + S²⁻ + Cu²⁺ + 2NO₃⁻ → 2NH₄⁺ + 2NO₃⁻ + CuS (s)
Then, the ions that occur on both sides of the overall ionic equation, are called spectator ions and they include 2NH₄⁺ and 2NO₃⁻.
These spectator ions can then cancel out, leaving us with the net ionic equation
S²⁻ + Cu²⁺ → CuS (s)
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