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ludmilkaskok [199]
3 years ago
14

If 1000g of water from Luboc River contains 0.0075g of Ca2+ ion, what is the concentration in ppm by mass of Ca2+ present in Lub

oc River?
Chemistry
1 answer:
marin [14]3 years ago
5 0

7.5 ppm

<h3>Further explanation</h3>

Given

1000 g of water

0.0075g of Ca²⁺ ion

Required

the concentration in ppm by mass of Ca²⁺

Solution

ppm = part per million

solvent = water ⇒ ppm = 1 mg/L(water density is 1 kg / L) or mg/kg

Convert g to mg of  Ca²⁺ ion :

0.0075 g = 7.5 x 10⁻³ g = 7.5 mg

Convert g to kg of water :

1000 g = 1 kg water

So the concentration of Ca²⁺ ion :

= 7.5 mg / 1 kg

= 7.5 ppm

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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

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Which of the following would have the largest pKa?
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then do the same for

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so CuO is the limiting reagent because there is less amount of it.

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