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3 years ago

A sample is found to contain 57.2 % N a H C O 3 NaHCOX3 by mass. What is the mass of NaHCO 3 in 4.25 g of the sample

Chemistry

1 answer:

liq [111]3 years ago

3 0

Answer:

The mass of N a H C O 3 present is 2.431 g

Explanation:

The sample contains 57.2 % N a H C O 3 by mass.

To find the mass of N a H C O 3 in the sample, we need to find what the equivalent of 57.2 %.

Mass of N a H C O 3 = Percentage Composition * Mass of sample

Mass of N a H C O 3 = 57.2 / 100 * 4.25

Mass of N a H C O 3 = 2.431 g

The mass of N a H C O 3 present is 2.431 g

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A 150 mL sample of hydrochloric acid (HCl) completely reacted with 60.0 mL of a 0.100 M NaOH solution. The equation for the reac

goblinko [34]

Answer:

0.040M

Explanation:

The following data were obtained from the question:

Volume of acid (Va) = 150mL

Volume of base (Vb) = 60mL

Molarity of base (Mb) = 0.1M

Molarity of acid (Ma) =..?

Next, the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Finally, we shall determine the concentration of the acid, HCl as follow:

MaVa / MbVb = nA/nB

Ma x 150 / 0.1 x 60 = 1/1

Cross multiply to express in linear form

Ma x 150 = 0.1 x 60

Divide both side by 150

Ma = 0.1 x 60 / 150

Ma = 0.04M

Therefore, the concentration of the acid is 0.04M

5 0

3 years ago

Read 2 more answers

Which of the following DOES NOT have 2 significant

Elanso [62]

Answer:

1.001

Explanation:

The Significant Figures are 1 0 0 1, This answer has 4 Significant figures, while the other three have only 2 significant figures

7 0

3 years ago

How many moles of NaCl are created as a result of decomposing 12 grams of NaClO3? (molar mass of Na=23, Cl=35, O=16)

Harman [31]

Answer: 0.113 moles of NaCl are created as a result of decomposing 12 grams of $NaClO_3$ .

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} NaClO_3=\frac{12g}{106.5g/mol}=0.113moles$

The balanced chemical equation for decomposition of $NaClO_3$ is:

$2NaClO_3\rightarrow 2NaCl+3O_2$

According to stoichiometry :

2 moles of $NaClO_3$ give = 2 moles of $NaCl$

Thus 0.113 moles of $NaClO_3$ give = $\frac{2}{2}\times 0.113=0.113moles$ of $NaCl$

Thus 0.113 moles of NaCl are created as a result of decomposing 12 grams of $NaClO_3$ .

8 0

3 years ago

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33

Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

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3 years ago

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UNO [17]

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3 years ago