Answer:
28g
Solution. Hence, 2 moles of nitrogen atoms weigh 28g.
Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
The reaction is:
Cl2 + 2 KBr --> 2 KCl + Br2
Moles of KCl is
n = m /M = 12 /74 = 0.16 mol
As, twice the moles of KCl is producing from 1 mol of chlorine
mole of Cl2 = 0.16 /2 = 0.08 mol
Mass of Cl2
m /70 = 0.08 = 5.6 g
Hence, 5.6 g mol Cl2 consumed to produce KCl
Answer:
P₂ = 1312.88 atm
Explanation:
Given data:
Initial temperature = 25°C
Initial pressure = 1250 atm
Final temperature = 40°C
Final pressure = ?
Solution:
Initial temperature = 25°C (25+273.15 = 298.15 K)
Final temperature = 40°C ( 40+273.15 = 313.15 k)
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
1250 atm / 298.15 K = P₂/313.15 K
P₂ = 1250 atm × 313.15 K / 298.15 K
P₂ = 391437.5 atm. K /298.15 K
P₂ = 1312.88 atm
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
