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3 years ago

Why did the can not become instantly crushed when plunged in the ice water when it was heated without any water inside?

1 answer:

3 0

Because since there's no water in the can, the cold water outside of it didn't have to equalize with the missing hot water that's missing from inside the can. if there was hot water inside the can then it was plunged inside the ice water, there it would have instantly been crushed.

Hope this helps

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Is H20 a balanced equation?

Nat2105 [25]

Answer:

no it's not

Explanation:

because H2+O2=H2O

so the balanced form is

H2 +O2=2H2O

8 0

3 years ago

a glass of ice at 0 degrees Celsius changes to a glass of water at 0 degrees Celsius what caused the ice to change to water

Tanya [424]

Thermal energy was absorbed.

6 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

PLEASEE ANSWERRR FASTTT

IceJOKER [234]

Answer:

last one

Explanation:

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3 years ago

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How much sodium bicarbonate to raise alkalinity in pool.

almond37 [142]

Answer:

A rule of thumb is that 1.5 lbs. of baking soda per 10,000 gallons of water will raise alkalinity by about 10 ppm. If your pool's pH is tested below 7.2, add 3-4 pounds of baking soda. If you're new to adding pool chemicals, start by adding only one-half or three-fourths of the recommended amount.

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2 years ago