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Afina-wow [57]
3 years ago
7

11111111111111111111

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

2222222222

Explanation:

mark me branliest of

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If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)
Ludmilka [50]

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.

4 0
3 years ago
Read the following chemical equations.
lora16 [44]

This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:

Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻  ⇒2K⁺O⁻²H⁺ + I₂⁰

Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻

Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.

In such a way, the correct choice is C.

Learn more:

  • (Redox reactions) brainly.com/question/13978139

8 0
2 years ago
Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling
Reil [10]

Answer:

The boiling  point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>

Explanation:

The formula for molal boiling Point elevation is :

\Delta T_{b} = iK_{b}m

\Delta T_{b} = elevation in boiling Point

K_{b} = Boiling point constant( ebullioscopic constant)

m = molality of the solution

<em>i =</em> Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

<em>i =</em> Van't Hoff Factor = 5

m = 8.5 m

K_{b} = 0.512 °C/m

Insert the values and calculate temperature change:

\Delta T_{b} = iK_{b}m

\Delta T_{b} = 5\times 0.512\times 8.5

\Delta T_{b} = 21.76 K

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

\Delta T_{b} = T_{b} - T_{b}_{pure}

T_{b}_{pure} = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0
3 years ago
A 10.0 g pat of butter raise the water level in a graduated cylinder by 11.6 mL. What is the density of the butter
Nina [5.8K]

- formula for density is mass divided by volume

therefore density of butter = 10.0g divided by 11.6ml = 0.8620689 g/cm³ ≈ 0.862 g/cm³ (3sf)

8 0
3 years ago
When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi
dexar [7]

Answer:

A. 266g/mol

Explanation:

A colligative property of matter is freezing point depression. The formula is:

ΔT = i×Kf×m <em>(1)</em>

Where:

ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g

Replacing in (1):

0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg

<em>0,0614 = mol of solute</em>.

As molar mass is defined as grams per mole of substance and the compound weights 16,0g:

16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>

I hope it helps!

3 0
3 years ago
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