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MrRissso [65]
2 years ago
12

How many moles are present in 2.3 x 1023 molecules of NH3?

Chemistry
1 answer:
avanturin [10]2 years ago
5 0

Answer:

<h2>0.38 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{2.3 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{2.3}{6.02}  \\  = 0.382059...

We have the final answer as

<h3>0.38 moles</h3>

Hope this helps you

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3 0
3 years ago
Stacy reads that a 50:50 mixture of methanol and water is best for keeping a car’s radiator from freezing when the temperature g
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1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.

2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.

3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.

4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.

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5 0
2 years ago
Read 2 more answers
An organic molecule is likely to contain all of these elements except: Answer C H O Ne N
eimsori [14]
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8 0
3 years ago
Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th
Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ =  C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0
2 years ago
For an aqueous solution of sodium chloride (NaCl) .Determine the molarity of 3.45L of a solution that contains 145g of sodium
worty [1.4K]
Molar mass NaCl = 58.44 g/mol

number of moles:

mass NaCl / molar mass

145 / 58.44 => 2.4811 moles of NaCl

Volume = 3.45 L

Therefore :

M = moles / volume in liters:

M = 2.4811 / 3.45

M = 0.719 mol/L⁻¹

hope this helps!
6 0
3 years ago
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