The thermodynamic law that explains the movement of heat energy during the refrigeration cycle is the first/zeroth/second.

2) 183 cg = kg 5 .00183 kg 0.0183 kg .183 kg 3) 0.25 kg = g

OlgaM077 [116]

Answer:

183 cg = 0.00183 kg

0.25 kg = 250 g

Explanation:

Use conversion factors. 1kg is equal to 1 x 10^5 cg (100000) and 1 kg is equal to 1 x 10^3 grams (1000 grams).

5 0

3 years ago

Out of 13 students how many saw themselves in the horoscope?

Andru [333]

Answer: Astrology is not astronomy! ... Astronomers and other scientists know that stars many light years* away have no effect on the ... Imagine a straight line drawn from Earth through the Sun and out into space way ... The 13 constellations in the zodiac. ... on the phases of the Moon), each month got a slice of the zodiac all to itself.

Explanation: Hope this helps

5 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

Need help with this question.... Next to each Formula, write the number of atoms of each element found in one unit of the compou

Andrew [12]

Answer:

The number of atoms present in one unit of the following compounds is:

a). Potassium Iodide , KI = 2

b).Sodium Sulfide, $Na_{2}S$ = 3

c). Silicon Dioxide , $SiO_{2}$ = 3

d). Carbonic Acid , $H_{2}CO_{3}$ = 6

Explanation:

Atomicity : It is defined as the number of atoms that are present in a given molecule/compound.

Atom : The smallest unit of matter is called atom. For e.g O is atom of oxygen but $O_{2}$ is not an atom , it is molecule of oxygen .

$O_{2}$ molecule has 2 atoms of Oxygen

Similarly Na , K , Fe are atoms but $O_{2}$ , $CO_{2}$ , $N_{2}$ , $H_{2}$ are molecules

a).Potassium Iodide

KI = 1 atom of K + 1 atom of I

Total atoms = 2

b) Sodium Sulfide

$Na_{2}S$ = 2 atoms of Na + 1 atom of S

Total atoms = 3

c) Silicon Dioxide

$SiO_{2}$ = 1 atom of Si + 2 atoms of O

Total atoms = 3

d) Carbonic Acid

$H_{2}CO_{3}$ = 2 atom of H + 1 atom of C + 3 atom of O

= 2+1+3

Total atoms = 6

7 0

3 years ago

Any solid that forms in a liquid during a chemical reaction is called a

Nimfa-mama [501]

it is called precipitant

7 0

3 years ago

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