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2 years ago

An aqueous salt solution is 15.0% mass sodium chloride. How many grams of salt are in 250.0 grams of this solution? Use correct

Chemistry

1 answer:

Shtirlitz [24]2 years ago

5 0

Answer:

37.5 g NaCl

Explanation:

Step 1: Given data

Concentration of NaCl: 15.0% m/m

Mass of the solution: 250.0 g

Step 2: Calculate how many grams of NaCl are in 250.0 g of solution

The concentration of NaCl is 15.0% by mass, that is, there are 15.0 g of NaCl every 100 g of solution.

250.0 g Solution × 15.0 g NaCl/100 g Solution = 37.5 g NaCl

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If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

kiruha [24]

The volume occupied by 2.00 moles of nitrogen gas at the same temperature and pressure will be

0.500 moles = 11.2 Liters
what about 2 moles =? liters

by cross multiplication

= 11.2 liters x 2moles/ 0.500 moles = 44.8 liters

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2 years ago

The first system of classification of organisms was developed by?

Margarita [4]

its the second one alright

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3 years ago

Read 2 more answers

Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c

inysia [295]

First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams

Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

Zinc mass = 0.37 x 954
Zinc mass = 352.98 g
Zinc moles = 352.98 / 65
Zinc moles = 5.43
Zinc atoms = 5.43 x 6.02 x 10²³
Zinc atoms = 3.27 x 10²⁴

Copper mass = 0.63 x 954
Copper mass = 601.02 g
Copper moles = 601.02 / 64
Copper moles = 9.39
Copper atoms = 9.39 x 6.02 x 10²³
Copper atoms = 5.56 x 10²⁴

3 0

2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

One calcium atom for every two iodine atoms chemical formula

borishaifa [10]

Is it multiple choice or do you just answer

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2 years ago