Answer:
a
Explanation:
All of the complexes will be colored except [Zn(en)3]2.
Fe, Ni, Cr, and Cu all form colored complexes because of their partially-filled d-orbitals and are generally referred to as transition elements. The d-orbital can split into higher and lower orbitals and electrons can jump from one level to the other. The jumping of electrons leads to light absorption and this characteristic impacts color to elements with partially-filled d-orbitals.
Zinc does not possess a partially-filled d-orbital and as such, its electrons cannot transition between levels, even though it belongs to the 'd-block' element.
The correct option is a.
How do you determine the uncertainty of a digital instrument?
Write a formula for the compound that forms from potassium and acetate. Express your answer as a chemical formula. Write a formula for the compound that forms from potassium and chromate.
Answer:
KCH₃COO
K₂CrO₄
Explanation:
Write a formula for the compound that forms from potassium and acetate.
Potassium has a charge +1 (K⁺) and acetate has a charge -1 (CH₃COO⁻). Thus, the neutral compound is KCH₃COO (potassium acetate).
Write a formula for the compound that forms from potassium and chromate.
Potassium has a charge +1 (K⁺) and chromate has a charge -2 (CrO₄²⁻). Thus, the neutral compound is K₂CrO₄ (potassium chromate).
How many moles of H2 can be formed if a 2.97 g sample of Mg reacts with excess HCl?
The number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles
To determine the number of moles of H₂ that can be formed if 2.97 g sample of Mg reacts with excess HCl,
First, we will write a balanced chemical equation for the reaction
Mg + 2HCl → MgCl₂ + H₂
From the reaction above,
It means that 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂
Now, we will determine the number of moles of Mg present in the given sample.
From the question,
Mass of Mg = 2.97 g
Using the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
Molar mass of Mg = 24.305 g/mol
∴ [tex]Number \ moles \ of \ Mg = \frac{2.97}{24.305}[/tex]
Number of moles of Mg in the sample = 0.122 moles
Since, 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂
∴ 0.122 moles of Mg will react with excess HCl to produce 0.122 moles of H₂
Hence, the number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles
Learn more here: https://brainly.com/question/18516616
In the diagram below, particles of the substance are moving from the liquid phase to the gas phi
they move from the gas phase to the liquid phase.
The gas and liquid are at
O equilibrium
a high vapor pressure.
a low vapor pressure.
zero vapor pressure.
Answer:
Equilibrium
Explanation:
Correct on edge 2020
Answer:
A - Equilibrium
Explanation:
For edge:)
a mixture of CaCO3 and CaO has a mass of 3.250g. After heating, the mass of the sample is reduced to 2.664g. What is the mass percent of CaCO3 in the mixture?
Answer:
40.92%
Explanation:
CaCo₃(s)-->CaO(s)+CO₂(g)
mass of Co2(g)= 1.33grams
mass of CaO(s)=(3.25-1.33)=1.92grams
mass % of CaCO₃ = [tex](\frac{1.33}{3.25} )[/tex]×100%
40.92%
How many grams of F are in 365 g CaF2 ?
Answer:
177 grams
Explanation:
moles of CaF₂ = [tex]\frac{mass}{molar mass}[/tex] = [tex]\frac{365}{78}[/tex] = 4.67949
moles of F-atom = 2×moles of CaF₂ = 2×4.67949 = 9.35898
(mol)×(molar mass)= 9.35898×19 = 177.82062
Therefore, 177 grams of F are in the mixture.
Determine the approximate density of a high-leaded brass that has a composition of 60.5 wt% Cu, 34.5 wt% Zn, and 5.0 wt% Pb. The densities of Cu, Zn, and Pb are 8.94, 7.13 and 11.35 g/cm3, respectively.
Answer:Approximate density of the high-leaded brass(alloy) =8.306g/cm³
Explanation:
The density of an alloy is its mass (100g) divided by its volume
Therefore we have that the alloy (high-leaded brass) has a composition of
60.5 wt% Cu with density 8.94g/cm3
34.5 wt% Zn, wth densty 7.13g/cm3
5.0 wt% Pb with densty g/cm3
The total volume of the alloy will be the mass / density of ts composition given as :
60.5gCu/DCu + 34.5gZn/DZn + 5.0gPb/Dpb
= 60.5/8.94 + 34.5/7.13+ 5.0/11.35
= 6.76 cm³ + 4.838 cm³+0.4405 cm³ = 12.0385cm³
Approximate density of the high-leaded brass(alloy) = 100g/ 12.0385cm³ =8.306g/cm³
You turn on the radio in the middle of a report about a planet. You don’t hear the planet’s name, but you hear that it is a large, gas giant with 66 confirmed moons and a Great Red Spot. Based on this information, which planet is being discussed?
Group of answer choices
Neptune
Uranus
Saturn
Jupiter
Answer:
Jupiter
Explanation:
Which of the following is a characteristic of living things.
1. They obtain and use material and energy.
2. They grow & develop 3They respond to their environment.
4. They are made up of complex & organized cells
5. All the above are characteristics of living things
Answer:
I would say all of the above
Explanation:
I say this because we use material and energy, we grow and develop, we are made up of cells, and we respond to the environment.
(NH4)3PO4 + CaBr2 = Ca3(PO4)2 + NH4Br
ionic equation
Answer:
2(NH4)3PO4 + 3CaBr2 → Ca3(PO4)2 + 6NH4Br
Explanation:
Tribasic Ammonium Phosphate + Calcium Bromide = Tricalcium Phosphate + Ammonium Bromide
Reaction Type
Double Displacement (Metathesis)
Reactants
Tribasic Ammonium Phosphate - (NH4)3PO4
Ammonium Phosphate
Molar Mass of H12N3O4P Oxidation State of H12N3O4P
Calcium Bromide - CaBr2
Kalziumbromid CaBr2 Calcium Bromide Hydrate
Molar Mass of Br2Ca Oxidation State of Br2Ca
Products
Tricalcium Phosphate - Ca3(PO4)2
Bone Phosphate Of Lime BCP TCP Tricalcium Orthophosphate Calcium Phosphate Tertiary Calcium Phosphate Calcium Orthophosphate Ca3(PO4)2
Molar Mass of Ca3O8P2 Oxidation State of Ca3O8P2
Ammonium Bromide - NH4Br
NH4Br
Molar Mass of BrH4N Oxidation State of BrH4N
why dose sodium have the same dot diagram as lithium
Answer:
They are in the same group on the periodic table indicating that they have the same number of valence electrons.
Explanation:
What are most membranes made up of?
layers of ducts
layers of epithelial tissue
layers of bursae
Answer:
layers of epithelial tissue
Answer:
It's B
Explanation:
The Jones family is expecting guests at any moment they spray room freshener in one corner of the kitchen to remove the strong foods smell The room spray quickly becomes evenly distributed throughout the room making the whole kitchen smell fresh what is happening in this system?
Answer: Entropy is increasing
Explanation:
How do you know that K, an alkali metal is very reactive?
a. It conducts heat
b. It conducts electricity
C. It is a soft and shiny metal
d. It contains one valence electron
Answer:
d. It contains one valence electron.
Since, potassium has one valence electron so it can easily donate one electron during chemical reaction.
A gas occupies 1.00 L at 273 K. What is the volume at 606 °C?
Answer:
3.22 L
Explanation:
0.00 C = 273 K
k = C +273
606 K + 273 K = 879 K
879/273 = 3.219
V2 = 3.22 L
Iodine – 125 emits gamma rays when it decays to Tellerium-125 and has a half-life of 60 days. If a 0.020 g pellet of iodine – 125 is implanted into a prostate gland, how much iodine – 125 remains there after one year
Answer:
0.0003125 g
Explanation:
Given data:
Half life of iodine-125 = 60 days
Mass of iodine pellet = 0.020 g
Mass of iodine remain after one year = ?
Solution:
one year = 365 days
Number of half lives = Time elapsed/ half life
Number of half lives = 365 days/ 60 days
Number of half lives = 6
At time zero = 0.020 g
At 1st half life = 0.020 g/2 = 0.01 g
At 2nd half life = 0.01 g / 2 = 0.005 g
At 3rd half life = 0.005 g/2 = 0.0025 g
At 4th half life = 0.0025 g / 2 = 0.00125 g
At 5tht half life = 0.00125 g / 2 = 0.000625 g
At 6th half life = 0.000625 g/ 2 = 0.0003125 g
How many moles are in 36 g of Be?
To make a new solution, 35.0 mL of a 12.0 M HCl stock solution is diluted to a volume of 1.20 L. What is the concentration of the new solution?
Answer:
m1v1=m2v2 change 35 ml to l = 0.035L
12*0.035 = M2*1.20
Solve for M2 = 0.35M
Explanation:
round 77.682517 to four figures
A gas was trapped in a container containing 10mL of air at a pressure of 1.0 atm. What will be the new volume,if the pressure is changed to 5 atm at the same temperature?
Answer:
The answer is 2 mLExplanation:
The volume of the gas can be found by using the formula for Boyle's law
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
We have
[tex]V_2 = \frac{10 \times 1}{5} = \frac{10}{5} \\ [/tex]
We have the final answer as
2 mLHope this helps you
At high temperatures, alkanes can undergo dehydrogenation to produce alkenes. For example: This reaction is used industrially to prepare ethylene while simultaneously serving as a source of hydrogen gas. Explain why dehydrogenation only works at high temperatures.
Answer:
Dehydrogenation of alkanes is endothermic
Explanation:
The dehydrogenation of alkanes is an endothermic reaction. The enthalpy change for the reaction is highly positive.
Recall that for an endothermic reaction, increase in temperature increases the rate of forward reaction.
As a result of that, the dehydrogenation of alkanes proceeds in the foward direction at elevated temperatures.
Compute the bond energy of the C-Cl bond using the reaction and data in question 10, the value of the bond energy for the C-H bond of 410 kJ/mol and the value of the H-H bond of 432 kJ/mol. The C-Cl bond energy in kJ/mol is:
Answer:
The bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
Explanation:
From the data in question 10:
The equation for the given reaction can be expressed as:
[tex]CH_{4(g)} + 2Cl_{(g)} \to CH_2Cl_2_{(g)}+H_2_{(g)}[/tex]
The enthalpy of formation of the compounds are as follows:
Substance [tex]\Delta \ H^0_f \ (kJ/mol)[/tex]
[tex]CH_{4(g)}[/tex] - 74.8
[tex]Cl_{(g)}[/tex] 120.9
[tex]CH_2Cl_{2(g)}[/tex] -95.8
From the data obtained above, the enthalpy of the reaction [tex]\Delta \ H ^0_{rxn}[/tex] can be computed as follows:
[tex]\Delta H^0_{rxn } = \sum ( \Delta H^0_f )_{products} - \sum ( \Delta H^0_f )_{reactants}[/tex]
[tex]\Delta H^0_{rxn } = [ 1 \ mol \times ( \Delta H^0_{CH_2Cl_{2(g)}}) + 1 \ mol \times ( \Delta H^0_{H_{2(g)}})] - [ 1 mol \times (\Delta H^0_{CH_4}_{(g)} +2 \ mol \times (\Delta \ H^0 _{Cl(g)})]}[/tex]
[tex]\Delta^0_{rxn } = [ 1 \ mol \times ( -95.8 \ kJ/mol) + 1 \ mol \times ( 0 \ kJ/mol] - [ 1 mol \times (-74.8 \ kJ/mol +2 \ mol \times (120.9 \ kJ/mol)]}[/tex]
[tex]\Delta H^0_{rxn } =-262.8 \ kJ[/tex]
However; since we knew [tex]\Delta H^0_{rxn }[/tex] to be -262.8 kJ
At standard conditions
Bond Bond Energy (kJ/mol)
C-H 410
H-H 432
The bond energy can be calculated by using the expression:
[tex]\Delta H^0_{rxn } = \sum \Delta H _{bond \ broken }- \sum \Delta H _{bond \ formed}[/tex]
[tex]-262. 8 KJ =[4 \times ( \Delta H_{CH})] - 1 [ (2 \times \Delta H_{CH} ) + ( 2 \times \Delta H_{C-Cl}) + (1 + \Delta H _{H-H})][/tex]
[tex]-262. 8 KJ =[4 \times ( 410 \ kJ/mol) ] - 1 [ (2 \times 410 \ kJ/mol ) + ( 2 \times \Delta H_{C-Cl}) + (1 +432 \ kJ/mol][/tex]
[tex]-262. 8 KJ = 388 \ kJ - ( 2 \times \Delta H_{C-Cl})[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 388 \ kJ +262. 8 KJ[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 650.8 \ KJ[/tex]
[tex]\Delta H_{C-Cl} = \dfrac{650.8 \ KJ}{2}[/tex]
[tex]\Delta H_{C-Cl} = 325.4 \ kJ[/tex]
Thus; the bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
4.60 g of 3-hexanol were obtained from 5.33 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles of hex-3-ene, , used in the experiment. (To avoid introducing rounding errors on intermediate calculations, enter your answer to four significant figures.)
Answer:
Explanation:
C₆H₁₂ + H₂O = C₆H₁₃OH
hexene hexanol
1 mole 1 mole
Mol weight of hexene = 84
Mol weight of hexanol = 102
5.33g of hexene = .06345 moles
4.6 g of hexenol = .0451 moles
percentage yield of hexanol
= .0451 x 100 / .06345
= 71.07 %
Moles of hexene used = .0451 moles
A 22.60 gram sample of chromium is heated in the presence of excess chlorine. A metal chloride is formed with a mass of 68.83 g. Determine the empirical formula of the metal chloride.
Answer:
CrCl₃.
Explanation:
From the question given above, the following data were obtained:
Mass of chromium (Cr) = 22.60 g.
Mass of metal chloride = 68.83 g.
Empirical formula =.?
Next, we shall determine the mass of chlorine (Cl) in the metal chloride. This can be obtained as follow:
Mass of chromium (Cr) = 22.60 g.
Mass of metal chloride = 68.83 g.
Mass of chlorine (Cl) =.?
Mass of chlorine (Cl) = (Mass of metal chloride) – (Mass of chromium)
Mass of chlorine (Cl) = 68.83 – 22.60
Mass of chlorine (Cl) = 46.23 g
Finally, we shall determine the empirical formula for the metal chloride as follow:
Cr = 22.60 g.
Cl = 46.23 g
Divide by their molar mass
Cr = 22.60 / 52 = 0.435
Cl = 46.23 / 35.5 = 1.302
Divide by the smallest
Cr = 0.435 /0.435 = 1
Cl = 1.302 /0.435 = 3
The empirical formula for the metal chloride is CrCl₃
how the charge of an element is used to determine the chemical formula if the elements have equal and unequal charges.
Answer:
The charges are typically used to determine the mole ratio and composition of the individual atoms of elements in the compound
Explanation:
Take for instance the following compounds:
[tex]NaOH[/tex] and [tex]H_{2} O[/tex]
Na has a charge of +1 and OH -1. equal charge means equal composition.H has charge of +1 while O has charge of -2. The charges are unequal, so the mole ratio for conversion would require multiplication by 2...which makes H two atoms but oxygen one atom in the compound.I hope this helps.
It would take more than 2___ wind turbines to replace coal power plants worldwide
Answer:
2,000,000
Explanation:
Answer:
yeah 2,000,000
Explanation:
The reference compound for naming D and L isomers of sugars is: A. glucose. B. ribose. C. glyceraldehyde. D. fructose. E. sucrose.
Answer:
C
Explanation:
The reference compound for naming D and L isomers of sugar is glyceraldehyde.
D and L stand for dexter (right) and laevus (left). The D and L system references molecules to that of glyceraldehyde which is chiral. By chirality, it means that glyceraldehyde molecules have the capacity to produce non-superposable mirror images. The images are labeled D and L isomers for 'right rotary' and 'left rotary' respectively.
This property allows some chemical manipulations to be performed on the molecules of glyceraldehyde without changing its configuration and as such, other compounds with similar ability are named by analogy to glyceraldehyde.
The correct option is C.
What element does this model represent?
A
Potassium
B
Sodium
C
Oxygen
D
Boron
the equilibrium constant, K, is greater than one. The stronger acid in this system is ___________ and the weaker base is ____________.
The question is incomplete; the complete question is;
For the following reaction:
H2CO3 + S2- ---------> HCO3- + HS-
The equilibrium constant is greater than one. The strongest acid in this system is ___________ and the weakest base is ____________.
A) H2CO3 and HS-
B) S2- and HS-
C) HCO3- and S2-
D) HS- and S2-
E) H2CO3 and HCO3-
Answer:
H2CO3 and HS-
Explanation:
According to the Brownstead- Lowry theory of acids and bases, an acid is a proton donor while a base is specie that accepts protons.
A strong acid has a greater tendency to donate protons. If we look at the reaction, H2CO3 has the greatest ability to donate protons hence it is the strongest acid.
On the other hand, HS- has the least ability to accept protons so it is the weakest base.
What molecules from food and air end up in the cells in the body?
PLEASE
Answer:
Molecules from food and molecules of oxygen move from the mouth and the nose to cells of the body through a series of blood vessels, including veins, arteries, and microscopically small blood vessels (capillaries), that extend throughout the body.Explanation:
Answer:
Molecules from food and air enter your body through your nose and mouth. These particles go down your throat and your lungs, and to your stomach. The blood also distributes these molecules.
Hope this helps!