In a laboratory experiment,john uses a mesh to separate soil particles from water.what technique of separation is he using?

A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the

bagirrra123 [75]

The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is
% by mass = (mass of the solute / mass of the solution)*100
= mass of solute / (mass of the solute + mass of the solvent)*100
= (27.7 g CaCl2 / 27.7g + 375g) * 100
= 6.88

5 0

3 years ago

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Which definition best describes global warming?

Marat540 [252]

Answer:

answer:-

a long-term increase in the Earth's average temperature

4 0

2 years ago

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What is the [OH-] of a substance that has a pH of 11?

Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0

2 years ago

Compare and contrast epithelial and muscle.

jekas [21]

Answer:

Epithelial tissues act as coverings, controlling the movement of materials across their surface. Connective tissue binds the various parts of the body together, providing support and protection. Muscle tissue allows the body to move and nervous tissues functions in communication.

Explanation:

6 0

3 years ago

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate

oee [108]

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

Initial temperature of water = 25.0°C

Final temperature of water = 29.0°C

Mass of water = 400.0 g

Mass of metal = 30.0 g

Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

$Q_{metal} = Q_{water}$

Mathematically, heat capacity or quantity of heat is given by the formula;

$Q = mc\theta$

<u>Where:</u>

Q is the heat capacity or quantity of heat.

m is the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Substituting the values into the formula, we have:

$30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}$

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

8 0

2 years ago