Answer:
see calculations in explanation
Explanation:
percent = part/total x 100%
part = ∑ atomic mass of element
hydrogen = 1.008 amu (atomic mass units)carbon = 12.011 amunitrogen = 14.007 amutotal = ∑ molecular mass of compound
= H amu + C amu + Namu
= 1.008 amu + 12.011 amu + 14.007 amu
= 27.026 amu
%H = (1.008amu/27.026amu)100% = 3.730%
%C = (12.011amu/27.026amu)100% = 44.442%
%N = (14.007amu/27.026amu)100% = 51.827%
Check results ∑%values = 100%
3.730% + 44.442% + 51.827% = 99.999% ≅ 100%
Which affects both deep water and surface currents in the ocean? the wind the temperature of the water the Coriolis effect None of the above
Answer:
the wind and the temperature of water both drive ocean currents
Answer:
B. Write 'T' for true or 'F' for false statements :
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What is the relationship between radius and diameter
Answer:
the radius is half the diameter
The radius of a circle is equal to half the diameter.
How much heat (in kJ) would need to be removed to cool 150.3 g of water from 25.60°C to -10.70°C?
Answer:
Q = -22.9 kJ
Explanation:
Given that,
Mass of water, m = 150.3 g
Water gets cool from 25.60°C to -10.70°C.
The specific heat of water, c = 4.2 J/g°C
The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=150.3\times 4.2 \times (-10.7-25.6)\\\\Q=-22914.738\\\\or\\\\Q=22.9\ kJ[/tex]
So, 22.9 kJ of heat is needed to be removed to cool.
True or false, rewrite it to make it true.
To balance a chemical equation, only the subscripts of a reactant and product can be changes
Answer:
false
Explanation:
Consider this chemical equation:
2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
Which substances dissolve in solution? Select all
that apply.
AgNO3
Cu
Cu(NO3)2
Ag
Answer is:AgNO3 and Cu(NO3)2
Answer:
the answer is very correct as AgNO3 and Cu(NO3)2
Answer:
A,C
Explanation:
the person above me is correct
Please help me
5. Where is potential energy decreasing?
А B C D
Answer:
D
Explanation:
I believe it is D. your kinetic energy would be at b. A, the cart would be going at a constant rate, because there is no hill or steep slope.
3.00 g of a fuel (MW = 650.0 g/mol) is combusted in a bomb calorimeter. The temperature of the calorimeter increases by 25.0 degrees Celsius. The heat capacity for the calorimeter has been determined to be 9.001kJ/C. What is the heat of reaction (in kJ/mol)?
Answer:
-4.88 × 10⁴ kJ/mol
Explanation:
Step 1: Calculate the heat gained by the calorimeter (Qcal)
We will use the following expression.
Qcal = C × ΔT
where,
C: heat capacity of the calorimeterΔT: temperature changeQcal = 9.001 kJ/°C × 25.0°C = 225 kJ
Step 2: Calculate the heat released by the fuel
According to the law of conservation of energy, the sum of the heat released by the fuel (Qf) and the heat absorbed by the calorimeter is zero.
Qf + Qcal = 0
Qf = -Qcal = -225 kJ
Step 3: Calculate the heat of reaction in kJ/mol
3.00 g of a fuel (MW = 650.0 g/mol) released 225 kJ of heat. The heat of reaction is:
-225 kJ/3.00 g × 650.0 g/mol = -4.88 × 10⁴ kJ/mol
What is the correct sequence of chemical transformations with the correct phases?
Answer:
1. Cu(s) 2. Cu(NO3)2 (aq) 3. Cu(OH)2 (s) 4. CuO (s) 5. CuSO4 (aq) 6. Cu(s)
Explanation:
Give the IUPAC name for the following alkanes and cycloalkanes
Answer:
1.3-bromo,2-chloro,4-floro hexane
3.bromocyclo pentane
Nickel carbonyl decomposes to form nickel and carbon monoxide, like this:
Ni(CO)4 → Ni + 4CO
At a certain temperature, a chemist finds that a 2.7L reaction vessel containing a mixture of nickel carbonyl, nickel, and carbon monoxide at equilibrium has the following composition:
Compound Amount
Ni(CO)4 0.597g
Ni 12.7g
CO 1.98g
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Answer: The value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
Moles of [tex]Ni(CO)_4[/tex] = [tex]\frac{0.597g}{170.7g/mol}=0.0035moles[/tex]
Moles of [tex]Ni[/tex] = [tex]\frac{12.7g}{58.7g/mol}=0.216moles[/tex]
Moles of [tex]CO[/tex] = [tex]\frac{1.98g}{28.01g/mol}=0.071moles[/tex]
Volume of solution = 2.7 L
Equilibrium concentration of [tex]Ni(CO)_4[/tex] = [tex]\frac{0.0035mol}{2.7L}=1.29\times 10^{-3}M[/tex]
Equilibrium concentration of [tex]Ni[/tex] = [tex]\frac{0.216mol}{2.7L}=0.08M[/tex]
Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.071mol}{2.7L}=0.026M[/tex]
The given balanced equilibrium reaction is,
[tex]Ni(CO)_4\rightleftharpoons Ni+4CO[/tex]
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[Ni]^1\times [CO]^4}{[Ni(CO_4]^1}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(0.08)^1\times (0.026)^4}{(1.29\times 10^{-3})^1}[/tex]
[tex]K_c=2.8\times 10^{-5}[/tex]
Thus the value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]
Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with liquid ammonia. Calculate the moles of ammonium phosphate produced by the reaction of of ammonia. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
The given question is incomplete. The complete question is:
Ammonium phosphate [tex](NH_4)_3PO_4[/tex] is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid [tex]H_3PO_4[/tex] with liquid ammonia. Calculate the moles of ammonium phosphate produced by the reaction of 1.7 mol of ammonia. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer: 0.57 moles of ammonium phosphate are produced by the reaction of 1.7 moles of liquid ammonia
Explanation:
The balanced chemical equation is:
[tex]H_3PO_4+3NH_4OH\rightarrow (NH_4)_3PO_4+3H_2O[/tex]
According to stoichiometry:
3 moles of liquid ammonia [tex](NH_4OH)[/tex] produces = 1 mole of ammonium phosphate [tex](NH_4)_3PO_4[/tex]
Thus 1.7 moles of liquid ammonia [tex](NH_4OH)[/tex] produces = [tex]\frac{1}{3}\times 1.7=0.57[/tex] mole of ammonium phosphate [tex](NH_4)_3PO_4[/tex]
Thus 0.57 moles of ammonium phosphate are produced by the reaction of 1.7 moles of liquid ammonia
Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 6.42 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 8.53 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide
Explanation:
[tex]6.42 \times \frac{1}{30} \times \frac{1}{1} \times \frac{46}{1} = 9.844[/tex]
[tex]8.53 \div 9.844 \times 100 = 86.65\%[/tex]
how many core electrons does magnesium ion have
Answer:
10 core electrons
Explanation:
Answer:
Magnesium has 10 core electrons
ASAP PLEASE!!! Approximately how much more of the solute potassium chlorate can be added to water if the temperature increases from 30°C to 65°C?
Answer: You have a solution of potassium chlorate containing 4 g at 65º C. How many additional grams of solute must be added to it, to make the solution saturated ? approximately 26 more grams [the solubility of potassium chlorate is ... water. The solution is allowed to cool. At what new temperature would crystals begin to.
Explanation:
Put the following experimental steps for the Acid-Base Equilibria (HCL and NaOH) in the correct order:
a. Perform a scout colorimetric titration. of diluted vinegar with NaOH.
b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometic titration of diluted vinegar with NaOH.
c. Perform a scout colorimetric titration of HCl with NaOH.
d. One pair performs a careful colorimetric titration of HCl with NaOH and the other pair performs a potentiometic titration of HCl with NaOH.
e. Dilute a solution of commercial vinegar to prepare it for titration.
Answer:
e. Dilute a solution of commercial vinegar to prepare it for titration.
a. Perform a scout colorimetric titration of diluted vinegar with NaOH.
b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometric titration of diluted vinegar with NaOH.
c. Perform a scout colorimetric titration of HCL with NaOH.
d. One pair performs a careful colorimetric titration of HCL with NaOH and the other pair performs a potentiometric titration of HCL with NaOH.
Explanation:
Titration is a process in which acid and base are introduced together to determine unknown concentration of acid or base. This is also known as neutralization reaction. The chemical process form salt and water as by products because acid molecules react with base molecules.
The experimental steps for the Acid-Base Equilibria in the correct order is e, a, b, c and d.
What is titration?Titration is a process in which we determine the concentration of any unknown acid and base by titrating them with the standard known concentration of base or acid.
In potentiometric titration, in the place of indicator potential difference is used across the analyte of electrolytic solution. And in colorimetric titration, absorbance of light is measured by the solution. Process of method which is involved in the titration is:
e. Dilute a solution of commercial vinegar to prepare it for titration.
a. Perform a scout colorimetric titration of diluted vinegar with NaOH.
b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometric titration of diluted vinegar with NaOH.
c. Perform a scout colorimetric titration of HCL with NaOH.
d. One pair performs a careful colorimetric titration of HCL with NaOH and the other pair performs a potentiometric titration of HCL with NaOH.
Hence correct sequence is e, a, b, c and d.
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What is the name of PbS2
Answer:
Lead sulfide.
Explanation:
describe one similarity and one difference between amylopectin and glycogen?
Answer: The main difference between amylopectin and glycogen is that amylopectin is an insoluble form whereas glycogen is a soluble form. The 1,4-alpha glycosidic bonds form the linear chain of both amylopectin and glycogen while 1,6-alpha glycosidic bonds form the branches.
Explanation:
does anyone know the answer to this
For the reaction shown, drsw the major(s) expected product.
Answer:
2
Explanation:
Which of the following would form a spontaneous reaction at 298 K?
O A. AH = 28 kJ/mol, AS = 0.109 kJ/(mol-K)
B. AH = 31 kJ/mol, AS = 0.093 kJ/(mol-K)
C. AH = 8.7 kJ/mol, AS = 0.008 kJ/(mol-K)
D. AH = 205 kJ/mol, AS = 0.210 kJ/(mol-K)
Answer:
A
∆H = 28 kJ/mol, ∆S = 0.109 kJ/(mol-k)
Explanation:
A P E X
The spontaneous reaction at 298 K is the one having
ΔH = 28 kJ/mol, ΔS = 0.109 kJ/(mol-K)
We know the formula of Gibb's free energy as-
ΔG = ΔH - TΔS
ΔG = 28 - (298 × 0.109)
ΔG = -4.48
If a reaction is spontaneous, change in Gibb's free energy is negative and change in entropy is positive.
Therefore, The spontaneous reaction at 298 K is the one having
ΔH = 28 kJ/mol, ΔS = 0.109 kJ/(mol-K)
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what is elimination reaction
What is the number of moles in 546 g strontium nitrite?
A. 0.388 mol
B. 2.58 mol
C. 329 mol
D. 3.04 mol
Gallium (Ga, 69.723 g/mol) is a metalloid obtained from its salts during the smelting of ores of other elements, like Zinc. has broad applicability in the electronics industry. It is also used as a safe replacement for mercury in thermometers as it melts at 29.8 °C and has a heat of fusion of 5.59 kJ/mol. What is the entropy change of 22 g of gallium in J/K as it melts when placed on a surface at 29.8°C?
Answer:
4.255 J/ K
Explanation:
Given data :
mass of Gallium = 16 gm
molar mass = 69.723 g/mol
hence no of moles = 16 / 69.723 = 0.23 moles --------- ( 1 )
Δh ( heat fusion ) = 5.59 KJ/mol
Temperature = 29.8°C = 302.8 k
Determine the entropy change of gallium
Δs ( entropy change ) = heat fusion / temperature
= ( 5.59 * 1000) J/mol / 302.8 = 18.46 J/k*mol
Hence entropy change of gallium
= Δs * no of Gallium moles
= 18.46 * 0.23 = 4.255 J/ k
I hope this is right haha
Combustion of hydrocarbons such as octane (C_8H_18) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.
1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid octane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.330 kg of octane are burned in air at a pressure of exactly 1 atm and a temperature of 10.0 ?C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.
Answer: 1. [tex]2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
2. The volume of carbon dioxide gas that is produced is 537 L
Explanation:
1. Combustion is defined as the type of chemical reaction where a hydrocarbon is combusted in the presence of oxygen to give carbon dioxide and water.
The balanced reaction for combustion of octane is:
[tex]2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
2.
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex] [tex]\text{Moles of octane}=\frac{0.330\times 1000g}{114.23g/mol}=2.89moles[/tex]
According to stoichiometry:
2 moles of octane produce = 16 moles of carbon dioxide
2.89 moles of octane produce = [tex]\frac{16}{2}\times 2.89=23.1moles[/tex] of carbon dioxide
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm
V = Volume of gas = ?
n = number of moles = 23.1
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]10.0^0C=(10+273)K=283K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{23.1\times 0.0821Latm/K mol\times 283K}{1atm}=537L[/tex]
Thus the volume of carbon dioxide gas that is produced is 537 L
How is this effect of dissolved
substances important in nature?
Explanation:
Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid. This is important to every living thing on earth. It means that wherever water goes, either through the air, the ground, or through our bodies, it takes along valuable chemicals, minerals, and nutrients.
if the amounts of greenhouse gases such as carbon dioxide and methane continue to increase in the atmosphere, what will happen to the average temperature of Earth?
If greenhouse gases in the atmosphere continues to increase, then the
average temperature of Earth will increase too.
Greenhouse gases such as carbon dioxide and methane deplete the ozone
layer which shields the earth from radiations from the Sun. An increase in
these gases will therefore allow for more radiations to get to the earth.
This therefore brings about an increase in the average temperature of the
earth.
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3. How many particles are present in 13.4 moles of carbon dioxide?
8.07. 1024 particles
04.49. 1022 particles
6.02. 1024 particles
6.02. 1023 particles
Answer:
[tex]\boxed {\boxed {\sf 8.07 *10^{24} \ particles}}[/tex]
Explanation:
To convert from moles to representative particles, Avogadro's number: 6.02*10²³ is used. This number tells us the amount of particles in 1 mole of a substance. We can use it as a ratio.
[tex]\frac {6.02*10^{23} \ particles}{1 \ mol \ CO_2}}[/tex]
Multiply by the given number of moles.
[tex]13.4 \ mol \ CO_2 *\frac {6.02*10^{23} \ particles}{1 \ mol \ CO_2}}[/tex]
The moles of carbon dioxide will cancel out.
[tex]13.4 *\frac {6.02*10^{23} \ particles}{1}}[/tex]
The denominator of 1 can be ignored and this becomes a simple multiplication problem.
[tex]13.4 * {6.02*10^{23} \ particles[/tex]
[tex]8.0668*10^{24} \ particles[/tex]
The original number of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 6 in the thousandth place tells us round the 6 to a 7.
[tex]8.07 *10^{24} \ particles[/tex]
13.4 moles of carbon dioxide has approximately 8.07*10²⁴ particles.
plz someone help me i dont get this and i need to finish this
Answer:
The height of A is 900
Explanation: Beside A, there is a number which explains what the height is so the answer is A.
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. AlCl3 and K3PO4
b. RbCO3 and NaCl
c. Na2CO3 and MnCl2
d. K2S and NH4Cl
e. CaCl2 and (NH4)2CO3
Answer:
See explanation
Explanation:
Let us see what happens when each solution is mixed;
a) AlCl3(aq) + K3PO4(aq) ------> 3KCl(aq) + AlPO4(s)
A precipitate is formed here
b) RbCO3(aq) + NaCl(aq) -------> This is an impossible reaction hence no solid precipitate is formed here
c) MnCl2(aq) + Na2CO3(aq) → 2NaCl(aq) + MnCO3(s)
A precipitate is formed.
d) K2S(aq) + 2NH4Cl(aq) ------> 2KCl(aq) + (NH4)2S(aq)
No solid precipitate is formed
e) CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s) + 2NH4Cl(aq)
A solid precipitate is formed
O2 + 2C0-2CO2
4.62 grams of oxygen and 2.98 grams of carbon monoxide are placed in a closed
reaction vessel and the mixture is ignited. Combustion occurs until one of the
gases is totally consumed.
1. Identify the limiting reagent and the excess reagent
2. Determine the amount of excess reagent.
3. Calculate the grams of carbon dioxide produced.
Answer: 1. [tex]CO[/tex] is the limiting reagent and [tex]O_2[/tex] is the excess reagent.
2. 2.91 g of excess reagent
3. 4.66 g of carbon dioxide produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} CO=\frac{2.98g}{28g/mol}=0.106moles[/tex]
[tex]\text{Moles of} O_2=\frac{4.62g}{32g/mol}=0.144moles[/tex]
[tex]O_2+2CO\rightarrow 2CO_2[/tex]
According to stoichiometry :
2 moles of [tex]CO[/tex] require = 1 mole of [tex]O_2[/tex]
Thus 0.106 moles of [tex]CO[/tex] will require=[tex]\frac{1}{2}\times 0.106=0.053moles[/tex] of [tex]O_2[/tex]
Thus [tex]CO[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles
Mass of excess reagent [tex]O_2=moles\times {\text {Molar mass}}=0.091moles\times 32g/mol=2.91g[/tex]
As 2 moles of [tex]CO[/tex] give = 2 moles of [tex]CO_2[/tex]
Thus 0.106 moles of [tex]CO[/tex] give =[tex]\frac{2}{2}\times 0.106=0.106moles[/tex] of [tex]CO_2[/tex]
Mass of [tex]CO_2=moles\times {\text {Molar mass}}=0.106moles\times 44g/mol=4.66g[/tex]