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Rama09 [41]
4 years ago
12

Rainwater will combine with which atmospheric gas in order to create carbonic acid?

Chemistry
1 answer:
mihalych1998 [28]4 years ago
4 0
Carbon Dioxide environmental gas so as to make carbonic corrosive that is consolidated with water. Carbon dioxide noticeable all around can disintegrate in rain water to frame carbonic corrosive, H2CO3. Carbonic corrosive is a powerless corrosive. It somewhat ionizes to frame hydrogen particles. The hydrogen particles from carbonic corrosive give common rain water a somewhat corrosive pH estimation of 5.6.
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Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4 What volume of 1.45 M Ca(OH)2 is needed to react with 25.0 moles of H2SO4?
Shkiper50 [21]

Answer:

We need 17.2 L of Ca(OH)2

Explanation:

Step 1: Data given

Concentration of Ca(OH)2 = 1.45 M

Moles of H2SO4 = 25.0 moles

Step 2: The balanced equation

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4

Step 3: Calculate moles Ca(OH)2

For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4

For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4

Step 4: Calculate volume of Ca(OH)2

Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2

Volume Ca(OH)2 = 25.0 moles / 1.45 M

Volume Ca(OH)2 = 17.2 L

We need 17.2 L of Ca(OH)2

3 0
3 years ago
What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?
Viefleur [7K]
Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³ 
y =  \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}
\boxed{y \approx 0.04\:mol}


Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

x = 40*0.04
\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark

Answer:
The mass is 1.6 grams
6 0
3 years ago
How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?
slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

n_{LiOH}=n_{HBr}

That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

Best regards.

8 0
3 years ago
Which model of an atom was molded after the solar system.
Crank
The answer is the bohr model was molded after the solar system
8 0
4 years ago
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NEED ASAP!!! Several samples of a solid was measured to have these masses and volumes listed in the
sammy [17]
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8 0
3 years ago
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