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Temka [501]
3 years ago
9

A mass of the crucible (g) 88.000g

Chemistry
1 answer:
nlexa [21]3 years ago
4 0
A)5.000g
b)mole=mass/molarmass
             molar mass of H2O=18g
             mass of water=5g
b)mole of water=5.000/18.000=0.2778g
c)mass of salt=2.342g
d)mole of salt=mass/molar mass
                     pure salt is NaCl
                     molar mass of NaCl=58.50g
 d)mole of salt=2.342/58.5=0.040mol
e)mole ratio=0.27:0.04

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Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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A chemist takes 50-gram sample of sulfur powder that has a melting point of 115.2 °C. What is the melting point of a 100-gram sa
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115.2 °C since melting point is an intensive property  

Step-by-step explanation:

The melting point of a substance does not depend on how much you have.

For example, the melting point of water is 0 °C, whether it is an ice cube from the refrigerator or in the frozen pond outside.

The freezing point of a substance is an <em>intensive property</em>.

Thus, the melting point of 100 g of sulfur is 115.2 °C because melting point in an intensive property.

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according to this exercise we have the following:

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Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.

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when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.

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