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schepotkina [342]
4 years ago
7

What mass of sodium phosphate is necessary in order to produce 5.00 g of calcium phosphate

Chemistry
1 answer:
stealth61 [152]4 years ago
6 0

Answer:

The answer to your question is the letter C

Explanation:

mass of Na₃PO₄ = ?

Ca₃(PO₄)₂ = 5 g

Balanced chemical reaction

                2Na₃PO₄  +  3CaCl₂  ⇒  6NaCl  + Ca₃(PO₄)₂

Process

1.- Calculate the molar mass of Sodium phosphate and calcium phosphate.

Na₃PO₄ = 2[(3 x 23) + (31 x 1) + (16 x 4)]

             = 2[69 + 31 + 64]

             = 2[164]

             = 328 g

Ca₃(PO₄)₂ = (3 x 40) + (31 x 2) + (8 x 16)

                 = 120 + 62 + 128

                 = 310 g

2.- Use proportions to find the mass of Na₃PO₄

              328 g of Na₃PO₄ -------------------- 310 g of Ca₃(PO₄)₂

                  x                        -------------------  5 g

                  x = (5 x 328) / 310

                  x = 1640 / 310

                  x = 5.29 g of Na₃PO₄              

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For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that w
Sindrei [870]

Answer:

75 mg

Explanation:

We can write the extraction formula as

x = m/[1 + (1/K)(Vaq/Vo)], where

x = mass extracted

m = total mass of solute

K = distribution coefficient

Vo = volume of organic layer

Vaq = volume of aqueous layer

Data:

m = 75 mg

K = 1.8

Vo = 0.90 mL

Vaq = 1.00 mL

Calculations:

For each extraction,

1 + (1/K)(Vaq/Vo) = 1  + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62  

x = m/1.62 = 0.618m

So, 61.8 % of the solute is extracted in each step.

In other words, 38.2 % of the solute remains.

Let r = the amount remaining after n extractions. Then  

r = m(0.382)^n.

If n = 7,

r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg

m = 75 - 0.088 = 75 mg

After seven extractions, 75 mg (99.999 %) of the solute will be extracted.

5 0
3 years ago
Can anybody check my answer?
anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:  

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

6 0
4 years ago
Lipid vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high con
NNADVOKAT [17]

Answer:

The question is not complete, the complete question should be "Lipids vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high concentration of solutes, the solution outside the vesicle is said to be Hypertonic. True or False"

The answer is True

Explanation:

This is because  it contains greater concentration of solutes on the outside of the cell than the increase.

In other words hypertonic solutions have more concentrate of solutions on the outside than the inside.

3 0
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