Answer:
Zn(s) + 2H₂O (l) → Zn(OH)2(aq) + H2(g)
Sn(s) + O2(g) →SnO2(s)
Cd(s) + Pb(NO3)2 (aq) →Cd(NO3)2 (aq) + Pb(s)
Cu (s) + HCl(aq) → can not occur
Explanation:
Metals reacts with liquid water to yield the corresponding metallic hydroxide and hydrogen gas as shown above; Zn(s) + 2H₂O (l) → Zn(OH)2(aq) + H2(g)
Tin reacts with oxygen to yield tin (IV) oxide. This is an oxidation reaction as shown; Sn(s) + O2(g) →SnO2(s)
The reaction between Cd and Pb(NO3)2 is a single replacement reaction. It is possible because Cd is above Pb in the electrochemical series. Hence the reaction occurs thus; Cd(s) + Pb(NO3)2 (aq) →Cd(NO3)2 (aq) + Pb(s)
Copper does not displace hydrogen from dilute acids (such as HCl) since copper is lower than hydrogen in the electrochemical series hence the reaction does not occur.
They're elements. Elements are the simplest form of any substance that cannot be broken down by any chemical methods, including electrolysis or even heat. Examples include oxygen, hydrogen etc.
These are 6 questions and 6 answers.
Question 1:
Answer: 33.7 atm
Explanation:
1) Data:
p=?
m = 1360.0 g N2O
V = 25.0 liter
T = 59.0°C
2) Formulas:
Ideal gas law: p V = n R T
n = mass in grams / molar mass
3) Solution
n = mass of N2O in grams / molar mass of N2O
molar mass of N2O = 2 * 14 g/mol + 16 g/mol = 44 g/mol
n = 1360.0 g / 44 g/mol = 30.9 mol
T = 59.0 + 273.15 K = 332.15 K
R = 0.0821 atm*liter / K*mol
=> p = nRT / V = 30.9 mol * 0.0821 [atm*liter / K * mol] * 332.15K / 25.0 liter = 33.7 atm
Answer: 33.7 atm
Question 2:
Answer: 204.5 liter
Explanaton:
1) Data:
m = 11.7 g of He
V = ?
p = 0.262 atm
T = - 50.0 °C
2) Formulas:
pV = nRT
n = mass in grams / atomic mass
3) Solution:
atomic mass of He = 4.00 g/mol
n = 11.7 g / 4.00 g/mol = 2.925 mol
T = - 50.0 + 273.15 K = 223.15 K
pV = nRT => V = nRT / p
V = 2.925 mol * 0.0821 [* liter / K*mol] *223.15K / 0.262 atm = 204.5 liter
Answer: 204.5 liter
Question 3.
Answer: 97.8 mol
Explanation:
1) Data:
Ethane
T = 15.0 °C
p = 100.0 kPa
V = 245.0 ml
n = ?
2) Formula
pV = nRT
3) Solution
pV = nRT => n = RT / pV
T = 15.0 + 273.15K = 288.15K
R = 8.314 liter * kPa / (mol*K)
n = 8.314 liter * kPa / (mol*K) * 288.15K / [100.0 kPa * 0.245 liter] = 97.8 mol
Answer: 97.8 mol
Question 4:
Answer: 113.67 K = - 159.48 °C
Explanation:
1) Data:
V = 629 ml of O2
p = 0.500 atm
n = 0.0337 moles
T = ?
2) Formula:
pV = nRT
3) Solution:
pV = nRT => T = pV / (nR)
T = 0.500 atm * 0.629 liter / (0.0337 mol * 0.0821 atm*liter/K*mol ) = 113.67 K
°C = T - 273.15 = - 159.48 °C
Question 5.
Answer: 5.61 g
Explanation:
1) Data:
V = 3.75 liter of NO
T = 19.0 °C
p = 1.10 atm
m = ?
2) Formulas
pV = nRT
mass = number of moles * molar mass
3) Solution:
pV = nRT => n = pV / (RT)
T = 19.0 + 273.15 K = 292.15 K
n = 1.10 atm * 3.75 liter / [ (0.0821 atm*liter / K*mol) * 292.15 K ] = 0.17 mol
molar mass of NO = 17.0 g/mol + 16.0 g/mol = 33.0 g/mol
mass = 0.17 mol * 33.0 g/mol = 5.61 g
Question 6:
Answer: 22.4 liter
Explanation:
1) Data:
STP
n = 1.00 mol
V = ?
Solution:
1) It is a notable result that 1 mol of gas at STP occupies a volume of 22.4 liter, so that is the answer.
2) You can calculate that from the formula pV = nRT
3) STP stands for stantard pressure and temperature. That is p = 1 atm and T = 0°C = 273.15 K
4) Clear V from the formula:
V = nRT / p = 1.00 mol * 0.0821 atm*liter / (K*mol) * 273.15 K / 1.00 atm = 22.4 liter
CnH2n since the equivalent unsaturation is equal with 1
Hello!
If a sample contains 21.2 g N, how many moles of N does it contain?
0.66 mol
1.51 mol
14.01 mol
297.01 mol
We have the following data:
m (mass) = 21.2 g
MM (Molar mass of Nitrogen) = 14 g / mol
n (number of moles) = ?
Formula:
Solving:
Answer:
1.51 mol
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I Hope this helps, greetings ... Dexteright02! =)
