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3 years ago

Determine the mass in grams of each nacl solution that contains 1.5g of nacl

Chemistry

1 answer:

Rama09 [41]3 years ago

7 0

Answer:- 3333 g of solution.

Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.

(i) 0.045% NaCl by mass

Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?

We could solve this using proportions...

(0.045/100) = (1.5/X)

0.045(X) = 1.5(100)

0.045X = 150

X = 150/0.045 = 3333

So, 1.5 g of NaCl is present in 3333 g of solution.

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2 years ago

There are 50 fish in the pond that have tags. You catch 25 fish, and 5 of them have tags. What is the best estimate of the numbe

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Explanation:

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A proportion is the equalization of two ratios.

Here you assume that the ratio of fish with tags to total fish that you catch is the same than the ratio of fish with tags to total fish in the pond.

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5 fish with tag / 25 fish = 50 fish with tag / x

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3 years ago

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A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is

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<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

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In this case, you know:

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3 years ago

When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?

irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

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2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

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3.- Calculate the limiting reactant

The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

1 mol of NH₄Cl ---------------- 1 mol of AgCl

0.00188 mol of NH₄Cl ------------- x

x = (0.00188 x 1) /1

x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

143.5 grams of AgCl -------------- 1 mol

x -------------------------------------------0.00188 moles of AgCl

x = (0.00188 x 143.5) / 1

x = 0.269 grams of AgCl

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