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hodyreva [135]
3 years ago
12

If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit

s of milliliters and liters? (a) How many mL will the box contain? (b) How many L will the box contain?
Chemistry
1 answer:
Blizzard [7]3 years ago
3 0

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

Hence, the volume of the box is 0.2175 L

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Answer:

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Explanation:

4 0
2 years ago
How many grams of Na2SO4 can be produced from 423.67 g of NaCl?
san4es73 [151]

Mass of  Na2SO4= 514.18 grams

<h3>Further explanation</h3>

Given

423.67 g of NaCl

Required

mass of  Na2SO4

Solution

Reaction

2NaCl + H2SO4 → Na2SO4 + 2HCl

mol NaCl :

= 423.67 g : 58.5 g/mol

= 7.24

From the equation, mol Na2SO4 :

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4 0
3 years ago
1 how many moles of sodium bicarbonate are needed to neutralize 0.9ml of sulphuric acid at stp​
Natali5045456 [20]

Answer:

8.0356 * 10^-5 moles of NaHCO3

Explanation:

Sulphuric acid = H2SO4

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The reaction between both compounds is given by;

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In the reactin above;

2 mol of NaHCO3 neutralizes 1 mol of H2SO4

At stp, 1 mol occupies 22.4 L;

1 mol = 22.4 L = 22400 mL

x mol = 0.9 mL

x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4

Since 2 mol = 1 mol from the equation;

x mol = 4.0178 * 10^-5

x mol = 2 * 4.0178 * 10^-5

x = 8.0356 * 10^-5 moles of NaHCO3

3 0
3 years ago
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first we need to find the empirical formula of nicotine

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C - 74.1 g - 74.1 g/12 g/mol = 6.17 mol

H - 8.6 g - 8.6 g / 1 g/mol = 8.6 mol

N - 17.3 g - 17.3 g / 14 g/mol = 1.23

divide all by the least number of moles

C - 6.17 / 1.23 = 5.01

H - 8.6 / 1.23 = 6.99

N - 1.23 / 1.23 = 1.00

when the atoms are rounded off to the nearest whole numbers

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H - 7

N - 1

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we have to find how many empirical units make up 1 molecule

number of empirical units = molecular mass / mass of 1 empirical unit

= 162.26 g/mol / 81 g = 2.00

there are 2 empirical units

molecular formula is - 2 (C₅H₇N)

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4 0
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mote1985 [20]

Answer:

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6 0
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