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schepotkina [342]
3 years ago
7

You have a balloon filled with hellum that has a volume of 4.91 cubic decimeters

Chemistry
2 answers:
mafiozo [28]3 years ago
8 0

Answer:

so the balloon can fill up with air. If the balloon pops, then the air goes away.

juin [17]3 years ago
5 0

Answer:

32

Explanation:

because the number of heliem is your aswer

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Charles's law states that
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The volume of a fixed mass of a gas is directly proportional to the temperature
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When matter changes into new of different substance
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4 years ago
Concerning the 10.0 mL of 0.50 M NaCl to 100 mL of solution: When a solution is diluted, does it change the number of moles diss
Mars2501 [29]

No, the dilution does not change the number of moles dissolved

Explanation:

We can see that,

The molarity of the solution was 0.50 M

The volume of the solution is 10 ml.

No of moles of the solute was= volume * concentration

                                                   = 10 X 10^-3* 0.50

                                                     = 5*10^-3 moles

When the solution is diluted from 10 ml to 100ml, the molarity or concentration  changes but number of moles remains constant.

The molarity of 100 ml solution will be

c=n/V

  = 5*10^-3*/100*10^-3

  = 0.05

  when the solution is diluted to 100ml from 10 ml molarity changes from 0.5M TO 0.05 M

8 0
3 years ago
Read 2 more answers
Calculate the molarity (M) of a solution containing 49.0 grams of H3PO4 in 500 mL of solution.
vlada-n [284]

Answer

The molarity (M) of the H3PO4 solution = 1.0 M

Explanation

Given:

Mass of H3PO4 = 49.0 grams

Volume of the solution = 500 mL = 500/1000 = 0.5 L

What to find:

The molarity (M) of the H3PO4 solution.

Step-by-step solution:

Step 1: Convert 49.0 grams H3PO4 to moles using the mole formula.

Mole=\frac{Mass}{Molar\text{ }mass}

The molar mass of H3PO4 = 97.994 g/mol

So,

Mole=\frac{49.0\text{ }g}{97.994\text{ }g\text{/}mol}=0.50\text{ }mol

Step 2: Calculate the molarity of the solution using the molarity formula.

Molarity=\frac{Mole}{Volume\text{ }in\text{ }L}

Putting mole = 0.50 mol and volume = 0.50L into the formula, we have;

Molarity=\frac{0.50mol}{0.50L}=1.0\text{ }M

The molarity (M) of the H3PO4 solution = 1.0 M

5 0
1 year ago
Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to
jenyasd209 [6]
<span>5.45 x 10^3 kg of sodium carbonate is needed to neutralize 5.04 kg of sulfuric acid. For this, I will assume you have pure H2SO4. So first, you need to calculate the molar mass of H2SO4 and Na2CO3. Lookup the atomic weights of all the elements involved. Atomic weight of Sodium = 22.989769 Atomic weight of Sulfur = 32.065 Atomic weight of Carbon = 12.0107 Atomic weight of Oxygen = 15.999 Atomic weight of Hydrogen = 1.00794 Molar mass of H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass of Na2CO3 = 2 * 22.989769 + 12.0107 + 3 * 15.999 = 105.987238 g/mol The balanced equation for the reaction of Na2CO3 with H2SO4 is Na2CO3 + H2SO4 => Na2SO4 + CO2 + H2O so for every mole of sulfuric acid to be neutralized, you need 1 mole of sodium carbonate. You can determine the number of moles of sulfuric acid you have and then calculate the mass of that many moles of sodium carbonate. But, there's an easier way. Just use the relative mass differences between sodium carbonate and sulfuric acid. So 105.987238 g/mol / 98.07688 g/mol = 1.080655 So that means for every kg of sulfuric acid, you need 1.080655 kg of sodium carbonate. Now do the multiplication. 5.04 x 10^3 kg * 1.080655 = 5.4465 x 10^3 kg. Since you only have 3 significant figures for your data, round the result to 3 significant figures, giving 5.45 x 10^3 kg</span>
3 0
3 years ago
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