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lesya [120]
3 years ago
15

How many phosphorus atoms are there in 1.75 mol of calcium phosphate, Ca3(PO4)2

Chemistry
1 answer:
elena-s [515]3 years ago
6 0

Answer:

It has 2 phosphorus atoms in 1 mol

so for 1.75 mol =2×1.75

Explanation:

need thanks and make me brainiest if it helps you

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HClO4 acid solution has a concentration of 5 Molarity. Calculate the concentration of this solution in
Nutka1998 [239]

Answer:

1. Percentage by weight = 0.5023 = 50.23 %

2. molar fraction =0.153

Explanation:

We know that

Molar mass of HClO4 = 100.46 g/mol

So the mass of 5 Moles= 5 x 100.46

       Mass (m)= 5 x 100.46 = 502.3 g

Lets assume that aqueous solution of HClO4  and the density of solution is equal to density of water.

Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.

We know that

Mass = density x volume

Mass of 1000 ml  solution = 1 x 1000 =1000     ( density = 1 gm/ml)

            m'=1000 g

1.

Percentage by weight = 502.3 /1000

Percentage by weight = 0.5023 = 50.23 %

2.

We know that

molar mass of water = 18 g/mol

mass of water in 1000 ml = 1000 - 502.3 g=497.9 g

So moles of water = 497.7 /18 mole

moles of water = 27.65 moles

So molar fraction = 5/(5+27.65)

molar fraction =0.153

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3 years ago
Atoms can gain lose or share electrons durning a chemical change why doesn't this cause the atoms identities to change?
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Rising greenhouse gases. Climate change. Rising energy costs. Declining fossil fuels reserves. With the arguments against fossil
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Answer:

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Crystallization from cooling magma describes one way that
Helen [10]
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6 0
3 years ago
Given 200ul of a 0.5mg/ml stock solution of BSA, how much do you pipet into a test tube so that you are adding 5ug of BSA to the
Trava [24]

<u>Answer: </u>10\mu L of volume needs to be pipetted out in the test tube.

<u>Explanation:</u>

We are given:

Mass of BSA to be formed = 5\mu g=0.005mg      (Conversion factor: 1mg=1000\mu g

Volume of stock solution = 200\mu L=0.2mL    (Conversion factor: 1mL=1000\mu L

It is also given that for the mass of BSA is 0.5 g, the volume used up is 1 mL

In order to have, 0.005 g, the volume of stock solution needed will be = \frac{1mL}{0.5g}\times 0.005g=0.01mL=10\mu L

Hence, 10\mu L of volume needs to be pipetted out in the test tube.

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