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irina1246 [14]
3 years ago
9

The diagram below shows part of the rock cycle.

Chemistry
2 answers:
snow_lady [41]3 years ago
5 0
The answer is igneous
Alika [10]3 years ago
5 0
I think the answer is igneous rocks.<span />
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Chemists recognize many different elements, such as gold, or oxygen, or carbon. Suppose you got some carbon, and started splitti
Otrada [13]

Answer:

An Atom.

Explanation:

An atom is the smallest particle of a chemical element that can exist. It the fundamental piece of matter. Therefore, splitting carbon into smaller pieces , the smallest piece that would still be called "carbon" would be an Atom of carbon. However, atom is also made of three subatomic particles called electrons, protons and neutrons.

6 0
3 years ago
What is an example of kinetic energy at work,1 a ball lodge in the tree, 1 a frisbee flaying in the air, 3 a car in the garage
AfilCa [17]

Answer:

a frisbee flaying in the air

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where;

  • K.E represents kinetic energy measured in Joules.
  • M represents mass measured in kilograms.
  • V represents velocity measured in metres per seconds square.

Hence, an example of kinetic energy at work is a frisbee flaying in the air because it would possess energy due to its motion in the air.

8 0
2 years ago
The symbols from the list and match them with the names of the
stich3 [128]
Be - Beryllium
S - sulfur
K - Potassium
C - Carbon
B - Boron
Ar - Argon
O - oxygen
Ne - Neon
Ca - Calcium
H - Hydrogen
3 0
2 years ago
1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0
3 years ago
If 185 mg of acetaminophen were obtained from a tablet containing 350 mg of acetamino- phen, what would be the weight percentage
Alex73 [517]

Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.

Amount of acetaminophen initially taken = 350 mg

Amount of acetaminophen obtained after recovery =185 mg

Weight percentage recovery =\frac{mass recovered}{mass originally taken}*100

                                                = \frac{185 mg}{350 mg}*100

                                               = 52.9%

8 0
3 years ago
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