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WARRIOR [948]
3 years ago
8

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp = 9.05×10−18?

Chemistry
1 answer:
saveliy_v [14]3 years ago
5 0
<span>M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9</span>
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C = 0.11 mol
V = 5.65 L
n = ???

n = C*V
n = 0.11 * 5.65
n = 0.622 mols

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x = 111 * 0.622
x = 69.0 grams CaCl2
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4 0
3 years ago
A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu
schepotkina [342]

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

0.030=\frac{x}{137}

x= 4.1 g

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3 0
3 years ago
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25
Svet_ta [14]

Answer:

a)  V_1=5ul

b)  v=20ul

Explanation:

From the question we are told that:

initial Concentration C_1=50mg/ml

Final Concentration C_2=10mg/ml

Final volume needs V_2 =25ul

Generally the equation for Volume is mathematically given by

C_1V_1=C_2V_2

V_1=\frac{C_1V_1}{C_2}

V_1=\frac{10*25}{50}

V_1=5ul

Therefore

The volume of buffer needed is

v=V_2-V_1\\\\v=25-5

v=20ul

3 0
3 years ago
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