Answer:
15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Explanation:
Given data:
Number of moles of ammonium sulfate formed = ?
Number of moles of ammonia = 30.0 mol
Solution:
Chemical equation:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄
Now we will compare the moles of ammonium sulfate with ammonia.
NH₃ : (NH₄)₂SO₄
2 : 1
30.0 : 1/2×30.0 = 15.0 mol
So 15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Answer:
Explanation: The molar mass and molecular weight of Cr2(SO3)3 is 344.182.
Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice
= latent heat of ice
Latent heat of ice = 334 KJ
⇒
= 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.
We can solve this problem by using Henry's law.
Henry's law states that the amount of dissolved gas is proportional to its partial pressure.

C is <span>the solubility of a gas.
</span><span>k is Henry's law constant.
</span><span>P is the partial pressure of the gas.
</span>We can calculate the constant from the first piece of information and then use Henry's law to calculate solubility in open drink.
0.12=4k
k=0.03
Now we can calculate the solubility in open drink.


Now we need to convert it to g/L. One mol of CO2 is 44.01<span>g.
</span>The final answer is: