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2 years ago

Apply Why does a box on the seat of a car

Chemistry

1 answer:

olga2289 [7]2 years ago

3 0

Answer:

im confused on what you mean

Explanation:

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1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca

neonofarm [45]

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

Solution for Part 1 :

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = $\frac{222}{100}\times 250=555g$

Therefore, the mass of silver nitrate is, 555 g

Solution for Part 2 :

Formula used : $S_1P_1=S_2P_2$ (at constant temperature)

where,

$S_1$ = initial solubility of methane gas = 0.026 g/L

$S_2$ = final solubility of methane gas

$P_1$ = initial pressure of methane gas = 1 atm

$P_2$ = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

$(0.026g/L)\times (1atm)=S_2\times (0.06atm)$

$S_2=0.433g/L$

Therefore, the solubility of the gas will be, 0.433 g/L

5 0

2 years ago

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×

Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0

3 years ago

What is the maximum number of p orbitals in any single energy level in an atom?

nordsb [41]

Answer is 3
s -1
p - 3
d - 5
f - 7

7 0

3 years ago

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

5 0

2 years ago

What is the name of this unstable isotope from number 9?

raketka [301]

Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.

6 0

2 years ago

Read 2 more answers