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3 years ago

Calculate the volume of 2.408x1024 molecules of water vapor at STP.

Chemistry

1 answer:

Tom [10]3 years ago

8 0

Answer:

Explanation: im not rlly sure you can go with what i said or you dont

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Please answer number12

densk [106]

Hydrogen maybe but I don’t know for sure

7 0

3 years ago

What is the molarity of a 10 L solution containing 5.0 moles of solute?

Setler79 [48]

Molarity = Moles of solute/ L(liters) of solution

So let's plug in the information.

5.0 moles/10L = 0.5 M

3 0

3 years ago

Read 2 more answers

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

3 0

3 years ago

KCI is a molecule. True or False

Monica [59]

Answer: duh

Explanation:

4 0

3 years ago

Read 2 more answers

what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

Hitman42 [59]

Answer:

9.85mL

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

Ca(OH)2 + 2HNO3 —> Ca(NO3)2 + 2H2O

From the balanced equation above,

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) =?

Mb (Molarity of base) = 0.155 M

Va (volume of the acid) = 28.8 mL

Ma (Molarity of acid) = 0.106 M

Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i.e the base) can be obtain as follow:

MaVa/MbVb = nA/nB

0.106 x 28.8 / 0.155 x Vb = 2/1

Cross multiply to express in linear form as shown below:

2 x 0.155 x Vb = 0.106 x 28.8

Divide both side by 2 x 0.155

Vb = (0.106 x 28.8) / (2 x 0.155)

Vb = 9.85mL

Therefore, the Volume of calcium hydroxide is 9.85mL

7 0

3 years ago