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vfiekz [6]
3 years ago
3

Match each scenario to the law that explains it. A balloon shrinks when it's taken outside in the winter. When the size of an ai

r chamber is increased, the air pressure decreases. A balloon expands when air is blown into it. A closed, flexible container expands when it's heated. Pressing on an inflated balloon decreases its size.
Chemistry
2 answers:
pishuonlain [190]3 years ago
7 0

Answer:

1) A balloon shrinks when it's taken outside in the winter.  Charles Law

2) When the size of an air chamber is increased, the air pressure decreases. Boyle's law

3) A balloon expands when air is blown into it. Avogadro's law

4) Pressing on an inflated balloon decreases its size. Boyle's law

5) A closed, flexible container expands when it's heated. Charles law

Explanation:

Given the statement of Charles law, the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Hence 1 and 5 are phenomena that represent Charles law.

On the other hand; Boyle's law states that the volume of a given mass of gas is inversely proportional to pressure at constant temperature. The scenario in 2 and 3 are phenomena that represents Boyle's law.

Avogadro's law states that the volume of a given mass of gas is directly proportional to the number of moles present. Hence the scenario in (3) above represents Avogadro's law.

CaHeK987 [17]3 years ago
6 0

Answer:

See the picture below

Explanation:

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9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of
Vlad [161]

Answer:

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

6 0
3 years ago
Suppose 'A' is a liquid aromatic compound with molecular weight 78 and burns with sooty flame. a.Give the name of the compound '
Oliga [24]

Aromatic compounds are compounds that contain carbon-carbon multiple bonds.

The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.

So,

(CH)n = 78

where n is the number of each atom present.

(12 +1)n = 78

n = 78/13

n = 6

The molecular formula of the compound is C6H6

When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.

When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.

brainly.com/question/24305135

3 0
3 years ago
I Need help ASAP!!!
AlexFokin [52]

Answer:

B. Temperature

Explanation:

A thermometer is a device used to measure temperature. Temperature is the average kinetic energy of a material's molecules or its average thermal energy. Heat is transferred to the thermometer, causing changes in its physical properties.

4 0
3 years ago
At 300K, the pressure inside a gas-filled container is 500 kPa. If the pressure inside decreases to 100 kPa, but the volume rema
alukav5142 [94]
PV = nRT

⇒P1/T1 = P2/T2
⇒500/300 = 100/T2
⇒ T2 = 100×300/500 = 60K

Answer is D
3 0
4 years ago
Read 2 more answers
Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has pre
brilliants [131]

Answer:

\large \boxed{\text{122 000 J}}

Explanation:

1. Calculate the energy needed

\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}}  = \text{ 29 120 cal}

2. Convert calories to joules

\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}

8 0
3 years ago
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