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MA_775_DIABLO [31]
2 years ago
7

If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)

?
Chemistry
1 answer:
Ludmilka [50]2 years ago
4 0

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln}    *100\%\\\\w/w\%=3.15\%

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

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A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a
Ymorist [56]

Answer:

Equilibrium constant Kc for the reaction will be 1.722

Explanation:

O2(g)+NO(g)→CO(g)+ NO2(g)

0.88    3.9        ---         ---

0.88x   3.9-x     x           x

GIVEN:

0.88X-X= 0.11

⇒ X=0.77

CO2(g)+NO(g) → CO(g) + NO2(g)

0.88       3.9         ---          ---

0.88-x    3.9-x      x            x

=              3.13       0.77      0.77

=0.11

Kc = \frac{[CO] *[NO2]} {[CO2]*[NO]}

     ={{0.77}×0.77÷{{0.11×3.13}}

     = 1.722

6 0
3 years ago
You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply
victus00 [196]

Answer:

a. 300 kg of Fertilizer

b. 225 kg of fertilizer

c.400 Kg of fertilizer

d.600 Kg of fertilizer

Explanation:

The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.

The percentages can be expressed as fractions as follows:

For nitrogen; 40/100 = 0.4

For phosphorus; 15/100 = 0.15

For potassium; 10/100 = 0.1

To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.

Quantity of fertilizer required to add to a hectare to supply;

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b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer

c. Phosphorus at  60 kg/ha = 60/0.15 = 400 Kg of fertilizer

d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer

5 0
2 years ago
When a lithium atom forms an Li+ ion, the lithium atom
gizmo_the_mogwai [7]
4, loses an electron. a plus sign indicates an atom is losing an electron while a minus sign indicates an atom is gaining an electron
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3 0
3 years ago
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What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a
Blababa [14]
First, we write the balanced equation for this reaction:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

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We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
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3 years ago
3. 79,789kJ = (293k) (377kJ)
KatRina [158]

the anwser would most likely be j=0

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3 years ago
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