4.0 moles were used. hope ive helped you! please rate brailiest. :)
Answer:
818.2 g.
Explanation:
- Molarity is the no. of moles of solute per 1.0 L of the solution.
<em>M = (no. of moles of NaCl)/(Volume of the solution (L))</em>
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M = 2.0 M.
no. of moles of NaCl = ??? mol,
Volume of the solution = 7.0 L.
∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)
∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.
- To find the mass of NaCl, we can use the relation:
<em>no. of moles of NaCl = mass/molar mass</em>
<em></em>
<em>∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) =</em> (14.0 mol)*(58.44 g/mol) = <em>818.2 g.</em>
The compound which is ionic is Kh
Kh is ionic because ionic compound is made between a metal and a non metal. K( potassium) is a metal while H (hydrogen) is a non metal. The bond between potassium and hydrogen form ionic bond whereby potassium loses one electron and hydrogen gain one electron to form an ionic compound.
Ca=40.07841 ,Nitrogen=14.006722 ,Oxygen15.99943
molar mass= (1xMW of Ca)+(2xMWof N)+(6x MW of oxygen)
Molar mass (molecular weight) of Ca(NO3)2 is 164.0884 g/mol
Answer:
a. 581.4 Pa
b. 3.33x10⁻⁴ mol/L
c. 3.49x10⁻⁴ mol/L
d. 0.015 g/L
Explanation:
a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:
pA = 0.9532*6.1
pA = 5.81452 mbar
pA = 5.814x10⁻³ bar
1 bar ----- 10000 Pa
5.814x10⁻³ bar--- pA
pA = 581.4 Pa
b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:
PV = nRT
Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.
n = PV/RT
n = (610*1)/(8.314*210)
n = 0.3494 mol
The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):
n = PV/RT
n = (581.4*0.9532)/(8.314*210)
n = 0.3174 mol
cA = n/V
cA = 0.3174/953.2
cA = 3.33x10⁻⁴ mol/L
c. c = ntotal/Vtotal
c = 0.3494/1000
c = 3.49x10⁻⁴ mol/L
d. The molar masses of the gases are:
CO₂: 44 g/mol
N₂: 28 g/mol
Ar: 40 g/mol
O₂: 32 g/mol
CO: 28 g/mol
The molar mass of the mixture is:
M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol
The mass concentration is the molar concentration multiplied by the molar mass:
3.49x10⁻⁴ mol/L * 43.36 g/mol
0.015 g/L
