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svlad2 [7]
3 years ago
6

Determine the mass (in grams) of NaCl in 294 grams of a 24.1% (m/m) NaCl solution. Be sure to report to the correct number of si

gnificant figures with no units.
I NEED HELP WITH THIS URGENTLY!!!!!
Chemistry
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

70.9 grams 3 sig figs

Explanation:

24.1% of 294 grams = 0.241(294 gram) = 70.854 grams ≅ 70.9 grams 3 sig figs

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How much heat energy is required to boil 66.7 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.
pochemuha

Answer:

91.7 kJ

Explanation:

Step 1: Given data

  • Mass of ammonia (m): 66.7 g
  • Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol

Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia

The molar mass of ammonia is 17.03 g/mol.

66.7 g × 1 mol/17.03 g = 3.92 mol

Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia

We will use the following expression.

Q = ΔH°vap × n

Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ

6 0
2 years ago
What is the mass of a steel cylinder that has a density of 75.0 g/ml and a volume of 12 ml?
liubo4ka [24]
Density can be calculated using the following rule:
Density = mass / volume

Therefore:
mass = density * volume

We are given that:
Density = 75 g/ml
volume = 12 ml

Substitute with these givens in the equation to get the mass as follows:
mass = density * volume
mass = 75 * 12
mass = 900 g
3 0
3 years ago
PLZ HELP ASAP WILL
amid [387]

Answer:

A) Mass = 32 g of KCl

Explanation:

Given data:

Mass of potassium chloride produced = ?

Mass of potassium chlorate = 52 g

Solution:

Chemical equation:

2KClO₃     →       2KCl + 3O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 52 g/ 122.55 g/mol

Number of moles = 0.424 mol

Now we will compare the moles of KClO₃ and KCl

             KClO₃      :       KCl

               2            :       2

          0.424         :       0.424

Mass of KCl:

Mass = number of moles × molar mass

Mass = 0.424 mol × 74.55 g/mol

Mass = 32 g

4 0
3 years ago
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