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3 years ago

Changes in which two characteristics can indicate a physical change?

Chemistry

2 answers:

liberstina [14]3 years ago

6 0

Answer:

A-Shape

C-state

Explanation:

reactivity and flammability are both indicators of chemicals changes

noname [10]3 years ago

5 0

Answer:

A. and C.

Explanation:

A P E X

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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago

When should you read the label on a chemical container?

atroni [7]

Always. You never know what kind of chemical you’re dealing with and how powerful it is.

6 0

3 years ago

If 60. liters of hydrogen gas at 546 K is cooled to 273 K at constant pressure, the new volume of the gas would be

Korvikt [17]

Answer:30 L

Explanation:

Initial Volume

Initial Temperature

546

Final Temperature

273

Final Vloume

Sol:-

Since the pressure is constant and the question is asking about temperature and volume, i.e,

⇒

⋅

273

546

⇒

Hence the new volume of the gas is

6 0

3 years ago

A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

Marrrta [24]

Answer:

$V_2=12.1L$

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$