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Nataly_w [17]
3 years ago
6

How would the volume of helium in a balloon be affected if the balloon was placed in a room where air temperature is lower?

Chemistry
2 answers:
Alexus [3.1K]3 years ago
8 0

Answer:

The volume will decrease.

Explanation:

According to Charles' Law, the volume of a gas is directly proportional to its temperature.

V = kT

Thus, if T decreases, V decreases.

Xelga [282]3 years ago
7 0

Explanation:

So,when balloon filled with helium gas was placed in lower temperature we will observe that the volume of the helium gas inside the helium will decrease due to which size of the balloon will also get decreased.

This is because volume is directly proportional to the temperature of the gas at constant pressure and number of moles. And this law is known as Charles law of

V\propto T    (At constant pressure and number of moles)

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How will the concentration of H+ and OH− ions change when a substance with a pH 3.2 is added to water? Both H+ and OH− will incr
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Also water H2O is made of H+ and OH- ions. so when an acidic substance is added to water the concentration of H+ ions increase.

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2 years ago
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Aee all sets of political views correct? i need long answer​
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Answer:

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8 0
2 years ago
Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0
3 years ago
Why do we have 2 high tides and 2 low tides a day?
Sunny_sXe [5.5K]
Whereas semidiurnal tides are observed at the equator at all times, most locations north or south of the equator experiencetwo unequal high tides and twounequal low tides per tidal day; this is called a mixed tide and the difference in height between successive high (or low) tides iscalled the diurnal inequality.
5 0
3 years ago
What is the concentration of FeCl3 in a solution prepared by dissolving 10.0 g of FeCl3 in enough water to make 275 mL of soluti
puteri [66]

Answer: The concentartion of solution will be 0.224 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute =\frac{\text{Given mass}}{\text{Molar mass}}=\frac{10.0g}{162.2g/mol}=0.0616moles

{V_s} = volume of solution in ml = 275 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{0.0616\times 1000}{275ml}=0.224mole/L

Therefore, the concentration of solution will be 0.224 M

7 0
3 years ago
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