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2 years ago

When a bottle of soft drink is shaken violently, CO2 gas escapes from solution and

Chemistry

1 answer:

jeka942 years ago

6 0

Answer:

The pressure decrease with the passage of time due to again dissolution of carbon-dioxide gas in the liquid solution.

Explanation:

In soft drinks, the carbon-dioxide gas is added in the drinks with high pressure because carbon-dioxide is a gas which is insoluble in soft drink at room temperature but soluble in the drinks at high pressure so when the pressure is removed from the soft drink, the carbon-dioxide gas releases in the air with the passage of time. But in close bottle , there is no place of escape so it again dissolve in the solution.

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_HNO3 + _Al(OH)3

bekas [8.4K]

Answer:

Option D. 3, 1, 3, 1

Explanation:

From the question given above,

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

The equation can be balance as follow:

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:

3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:

3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃

Now, the equation is balanced.

Thus, the coefficients are 3, 1, 3, 1

7 0

3 years ago

How many moles of salt are in 13.8 g of sodium chloride?

Hoochie [10]

Answer: 0.24 moles

Explanation:

Molecular Mass of NaCl (23 + 35.5) = 58.5g

58.5g of Sodium Chloride -------> 1 mole of NaCl

∴ 13.8g of Sodium Chloride ------> 1 ÷58.5 x 13.8 = 0.2358974 ≈ 0.24moles

3 0

3 years ago

Read 2 more answers

HOW DO YOU GET ANY THING YOU WANT PLS

r-ruslan [8.4K]

Answer:

enough money can buy you anything

6 0

3 years ago

Read 2 more answers

How many molecules of nitrogen gas are found in 0. 045 L of nitrogen gas at STP?

stepan [7]

Answer:

See below

Explanation:

.045 liter / 22.4 l / mole * 6.022 x 10^23 molecules/mole * 2 atoms/molecule =

( * 2 becuase nitrogen gas is diatomic)

7 0

1 year ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago