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3 years ago

How will an increase in wind speed affect soil erosion?

Chemistry

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

A. Soil erosion will increase.

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What are three main classifications of the elements

neonofarm [45]

Answer:

Metals, non metals and metalloid

Explanation:

4 0

4 years ago

Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56 g of NaCl in 164.2 g of H2O. Express your answer to

denis-greek [22]

Answer:

"23.896%" is the right answer.

Explanation:

The given values are:

Mass of NaCl,

= 51.56 g

Mass of H₂O,

= 165.6 g

As we know,

⇒ Mass of solution = $Mass \ of \ (NaCl+H_2O)$

= $51.56+164.2$

= $215.76 \ g$

hence,

⇒ $Mass \ percent =\frac{Mass \ of \ NaCl}{Mass \ of \ solution}\times 100$

$=\frac{51.56}{215.76}\times 100$

$=23.896 \ percent$

4 0

3 years ago

Determine the molarity for the fixer Sodium carbonate decahydrate (I dissolved 127.62 grams of sodium carbonate for every 4 lite

grin007 [14]

Answer:

0.112 M.

Explanation:

Molarity is the no. of moles of solute in a 1.0 L of the solution.

M = n/V.

M = (mass/molar mass)solute x (1000/V of the solution).

mass = 127.62 g.

molar mass = 286.138 g/mol.

V of the solution = 4.0 L = 4000.0 mL.

∴ M = (mass/molar mass)solute x (1000/V of the solution) = (127.62 g / 286.138 g/mol) x (1000 / 4000.0 mL) = 0.1115 M ≅ 0.112 M.

4 0

4 years ago

What is the molarity of the 8.76 g of Na2S dissolved in 0.500 L of solution?

AlexFokin [52]

Answer:

.224 M

Explanation:

To begin, we need to find the molar mass of Na2S

Na=22.99g, S=78.05g

2(22.99)+1(32.07)=78.05

To find Molarity, we use the equation M=moles/Liters of solution

We are given grams, so we must divide by the molar mass of Na2S to find moles.

8.76g Na2S / 78.05g/mol Na2S= .112 moles

Now we use our Molarity equation:

.112 moles Na2S / .500L of solution = .224 M

6 0

2 years ago

A sample of cobalt has a mass of 27 g and a density of 9 g/cm^3. What would be its volume?

8_murik_8 [283]

Given ,

Mass of sample of cobalt = 27 g

density of sample of cobalt = 9g/cm^3

We know that ,

Density = mass of sample/volume of sample

From that relation ,

We can deduce the following as

Volume = mass of sample/density of sample

Hence , required volume of sample of cobalt = 27 g /9 g/cm^3 = 3 cm^3

The volume is $\fbox{3cm cube}$

3 0

3 years ago