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Alex Ar [27]
3 years ago
8

Josh is making homemade lollipops. the recipe calls for dissolving 400 g sugar in 100 g of water. his friend, alex, said that wo

uld never work because sugar's solubility is 204 g per 100 g of water at 20°c. despite alex's doubts, josh successfully dissolves all 400 g of the sugar. what did josh do that allowed him to dissolve that much sugar in only 100 g of water? josh used sugar cubes instead of granulated sugar. josh used a pan with a large surface area. josh stirred the solution. josh heated the water.
Chemistry
2 answers:
myrzilka [38]3 years ago
7 0
D.) josh heated the water
Juli2301 [7.4K]3 years ago
3 0

The correct answer is "josh heated the water."

The solubility of sugar is 204 g per 100 g of water only at 20°C. The solubility of sugar increases with increase in temperature. At high temperatures the kinetic energy of water molecule increases which can now effectively break apart the sugar molecules that are held together by intermolecular forces of attraction.

Thus 400 g of the sugar will dissolve in water at a temperature higher than  20°C.

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Equation: CH4 + 202 CO2 + 2H20
jeka94

Answer:

2:1

1.2 × 10² g

Explanation:

Step 1: Write the balanced combustion equation

CH₄ + 2 O₂ ⇒ CO₂ + 2 H₂O

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of O₂ to CH₄ is 2:1.

Step 3: Calculate the moles of CH₄ required to react with 15 moles of O₂

We will use the previously established molar ratio.

15 mol O₂   1 mol CH₄/2 mol O₂ = 7.5 mol CH₄

Step 4: Calculate the mass corresponding to 7.5 moles of CH₄

The molar mass of CH₄ is 16.04 g/mol.

7.5 mol × 16.04 g/mol = 1.2 × 10² g

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3 years ago
Consider the statement, “In a chemical reaction, atoms are neither created nor destroyed, only rearranged.” Why is this a statem
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7 0
3 years ago
A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
statuscvo [17]

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

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