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Jlenok [28]
2 years ago
11

What will determine the number of moles of hydronium in an aqueous solution of a strong monoprotic acid?.

Chemistry
1 answer:
marin [14]2 years ago
3 0

Answer:

What will determine the number of moles of hydronium in an aqueous solution of a strong monoprotic acid? The amount of acid that was added.

Explanation:

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Minchanka [31]
4) the ice has to be heated up to above 0c first, as it is continuing to cool the water back toward 0c
5) 0.0338 kcal
(33.8/1000)
6) 1.2506 kcal
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5 0
3 years ago
Calculate the mass of Na
Alexus [3.1K]

Answer:

22.98977

Explanation:

Na is actually sodium and its atomic number is 11.

I'm not sure if this is multiple choice or a question where you have to answer in writing but here is the mass of Na 22.98977.

Hope this helps

3 0
3 years ago
How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0℃.
Ivanshal [37]

Answer:

There will react 30.9 milligrams of magnesium

Explanation:

Step 1: Data given

Volume of hydrogen = 31.2 mL

Pressure = 754 torr = 754/760 = 0.992 atm

Temperature = 25.0 °C = 298 Kelvin

Step 2: The balanced equation

Mg + 2HCl → MgCl2 + H2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 0.992 atm

⇒with V = the volume of H2 = 31.2 mL = 0.0312 L

⇒ with n = the moles of H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 25.0 °C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.992*0.0312)/(0.08206*298)

n = 0.00127 moles

Step 4: Calculate moles of Mg

For 1 mol of Mg we need 2 moles of HCl to produce 1 mol of MgCl2 and 1 mol of H2

For 0.00127 moles of H2 we need 0.00127 moles of Mg

Step 5: Calculate mass of Mg

Mass of Mg = moles of Mg * molar mass of Mg

Mass of Mg = 0.00127 moles * 24.3 g/mol

Mass of Mg = 0.0309 grams = 30.9 mg of Mg

There will react 30.9 milligrams of magnesium

3 0
3 years ago
What type of animal might become preserved in amber
goldfiish [28.3K]
Insect's or Bugs preserved in Amber
6 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

6 0
3 years ago
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