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3 years ago

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Calculate the solubility (in M units) of ammonia gas in water at 298 K and a partial pressure of 7.00 bar . The Henry’s law cons

tant for ammonia gas at 298 K is 58.0 M/atm and 1 bar=0.9869 atm.

1 answer:

Juliette [100K]3 years ago

6 0

Explanation:

just for friendship kvp-hxns-jst

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Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate.

Reika [66]

Answer:

0.0428 M

Explanation:

Because we're asked to calculate the molarity of nickel(II) cation, we need to <u>determine all sources for that species</u>, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).

We use the molecular weight of NiBr₂ to calculate the moles of Ni:

1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) = 8.55x10⁻³ mol Ni⁺²

Then we <u>divide the moles by the volume in order to calculate the concentration</u>:

8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M

7 0

3 years ago

Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of

Neko [114]

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles$

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

$\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles$

Finally, the mass of SiH₄ is:

$m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g$

Therefore, 10.60 grams of silane gas are formed.

I hope it helps you!

3 0

3 years ago

If 13.7 grams of manganese oxide reacts with excess hydrogen chloride gas, how many grams of water are formed?

Lerok [7]

Answer:

5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.

Explanation:

EQUATION FOR THE REACTION

Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O

From the balanced reaction between manganese oxide and hydrogen chloride gas;

1 mole of MnO2 reacts to form 2 mole of water

At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:

(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water

(55 + 32) g of MnO2 reacts to form 2 * 18 g of water

87 g of MnO2 reacts to form 36 g of water

If 13.7 g of MnO2 were to be used?

87 g of MnO2 = 36 g of H2O

13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water

= 493.2 / 87 g of water

Mass of water = 5.669 g of water

Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.

3 0

3 years ago

Which is heavier? A pound of cotton candy or a pound of marbles

Ugo [173]

Answer: both are equal

They weighed the same because they both a pound

6 0

3 years ago

Read 2 more answers

The capacity of a muscle to continue contracting over a period of time without fatigue is called:

siniylev [52]

<span>B. muscular endurance

</span>The capacity of a muscle to continue contracting over a period of time without fatigue is called muscular endurance.
NOT:
A. muscular strength.
<span>C. cardiovascular endurance. </span>
<span>D. power.</span>

6 0

4 years ago