The answer is: 8.14·10⁶³ moles of lithium are present.
N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.
n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.
n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.
n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate
In one molecule of lithium sulfate, there are two atoms of lithium.
n(Li₂SO₄) : n(Li) = 1 : 2.
n(Li) = 2 · 4.07·10⁶³ mol.
n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.
Answer:
The molar mass of the gas is 36.25 g/mol.
Explanation:
- To solve this problem, we can use the mathematical relation:
ν = 
Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,
R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,
T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,
M is the molar mass of the gas in <em>(Kg/mol)</em>.
ν = 
(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,
<em>by squaring the two sides:</em>
(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,
∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.
<em>∴ The molar mass of the gas is 36.25 g/mol.</em>
Answer:
If fixed amount of gas is heated then the volume will increase because the heat will cause the molecules of gas to move freely and increase the kinetic energy.
Explanation:
If fixed amount of gas is heated then the volume will increase because the heat will cause the molecules of gas to move freely and increase the kinetic energy.
According to the Charle's law
The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.
Mathematical expression;
V ∝ T
V = kT
V/T = k
if volume is changed from V1 to V2 and temperature change from T1 to T2 then,
V1/T1 = k V2/T2= k
V1/T1 = V2/T2
Answer:

Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
![K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%200.0150x%20%3D%203.44%20%5Ctimes%2010%5E%7B-7%7D%5C%5Cx%20%3D%20%5Cdfrac%7B3.44%20%5Ctimes%2010%5E%7B-7%7D%7D%7B0.0150%7D%20%3D%20%5Cmathbf%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
![K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%20x%20%5Ctimes%202.293%20%5Ctimes%2010%5E%7B-5%7D%20%3D%202.53%20%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5Cx%20%3D%20%5Cdfrac%7B2.53%20%5Ctimes%2010%5E%7B-8%7D%7D%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%20%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Pb%24%5E%7B2%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
Answer:
29.98kg
Explanation:
12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg