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Tanya [424]
2 years ago
7

You have to prepare 100.0 mL of a 0.100 M solution of sodium carbonate. You have a concentrated solution of sodium carbonate tha

t you can take a small amount from and transfer to a 100.0 mL volumetric flask, and then dilute accordingly. If the concentrated solution has a concentration of 1.45 M, calculate the volume that you need to transfer to the volumetric flask.
Chemistry
1 answer:
Georgia [21]2 years ago
4 0

Answer:

6.9 ml of concentrate

Explanation:

100 ml   of .1 M   will require .01 moles

from a 1.45 M solution,  .01 mole would be

  .01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate   then dilute to 100 ml

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What is one way a pathogen can enter the body?
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I’m pretty sure it’s Nose
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3 years ago
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Which statement summarizes the difference between mass and weight?
alexgriva [62]

Answer:

Mass is the amount of matter in an object.

Weight is how much an object weighs.

Hope this helps!

6 0
3 years ago
If a sample of magnesium with a mass of 35.0 grams reacts with 35.0 grams of oxygen how much magnesium oxide will be produced?​
slamgirl [31]

Answer:

58.0 g of MgO

Explanation:

in a perfect world, 70 g, however we don't live in a perfect world

The equation of reaction

2Mg + O₂ --> 2MgO

first find which element is limiting:

35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO

35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO

This means Mg is the limiting factor, so you will be using this moles to find grams of MgO

1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO

8 0
3 years ago
1 mole solution of NaCl contains
Mamont248 [21]

Answer:

If you dissolve 58.44g of NaCl in a final volume of 1 liter, you have made a 1M NaCl solution, a 1 molar solution.

Explanation:

7 0
3 years ago
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
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