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3 years ago

10

How many grams of sodium bicarbonate are in 1.8 x 1023 formula units?NaHCO3

1 answer:

Makovka662 [10]3 years ago

8 0

Answer:

Mass = 25.2 g

Explanation:

Given data:

Number of formula units of NaHCO₃ = 1.8×10²³

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ formula unit

1.8×10²³formula unit × 1mol / 6.022 × 10²³ formula unit

0.3 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.3 mol × 84 g/mol

Mass = 25.2 g

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4 years ago

What is the molarity of a solution made from 325.4 g of AlCl3 (MM 133.34 g/mol) to make 500.0 mL?

NISA [10]

Answer : The molarity of a solution is, 4.88 M

Explanation : Given,

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Formula used :

$\text{Molarity}=\frac{\text{Mass of }AlCl_3\times 1000}{\text{Molar mass of }AlCl_3\times \text{Volume of solution (in mL)}}$

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3 years ago

Read 2 more answers

Oil (specific gravity of 0.80 and a viscosity of 0.000042 lbf/ft2) at a temperature of 80 F flows through two separate pipes 10

Elenna [48]

Answer:

The velocity of flow in 10in pipe is 4.16 ft/s.

Explanation:

Given that

Specific gravity = 0.8

Viscosity =0.00042 $lbf/ft^2$

For pipe 1

$V_1=6 ft/s,d_1=12\ in$

For pipe 1

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If we assume that flow in the both pipe is laminar

For laminar flow through circular pipe

$\dfrac{\Delta P}{L}=\dfrac{32V\mu }{d^2}$

So same pressure drop we can say that

$\dfrac{V_1 }{d^2_1}=\dfrac{V_2}{d^2_2}$

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$V_2=4.16 ft/s$

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4 years ago

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Vadim26 [7]

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4 years ago

If a centiliter is one hundredth of a liter (0.01 L), how many centiliters are there in one liter?

victus00 [196]

Answer:

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Explanation:

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