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inessss [21]
3 years ago
9

6

Chemistry
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, <u><em>the  pressure of the gas decreased by a factor of 3.</em></u>

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or

P * V = k

To obtain the proportionality factor k you must make the quotient:

k=\frac{V2}{V1} =\frac{6 L}{2 L}

k= 3

This means that <u><em>if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the  pressure of the gas decreased by a factor of 3.</em></u>

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What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza
puteri [66]

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

5 0
3 years ago
Explain the neutralization of vitamin c with sodium hydrogen carbonate?
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Scientists have known the consequences of mixing sodium bicarbonate and vitamin C ever since 1936, when a study on the subject was published in the "Journal of Nutrition." The authors of this study measured the amount of vitamin C recovered from the urine of people who drank a fixed amount of orange juice. The authors determined that the amount of vitamin C excreted was decreased by administration of sodium bicarbonate. Followup studies in the 1940s showed that this effect was due to the neutralization of the vitamin C by the sodium bicarbonate.
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3 years ago
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The clouds of Venus consist mainly of ______. Group of answer choices carbon monoxide droplets of water vapor hydrogen and heliu
Zanzabum

Answer:

droplets of sulfuric acid

Explanation:

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3 years ago
Identify the balanced equation for the following reaction:<br><br> SO2(g) + O2(g) → SO3(g)
sergiy2304 [10]

Answer:  The balanced equation for the given reaction is

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 1
  • O = 4

Number of atoms on product side are as follows.

  • S = 1
  • O = 3

To balance this equation, multiply SO_{2} by 2 on reactant side and multiply SO_{3} by 2. Hence, the equation will be re-written as follows.

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 2
  • O = 6

Number of atoms on product side are as follows.

  • S = 2
  • O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is 2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

3 0
3 years ago
An excess of Ba(No3)2 reacts with 250ml of H2SO4 solution to give 0.55g of BaSo4.determine The concentration in moles per litre
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m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
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