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3 years ago

PLEASE ANSWER

Chemistry

1 answer:

Natasha2012 [34]3 years ago

7 0

Refraction explains why a stone looks larger inside of a fish bowl than out of the fish bowl?

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Where does most the candles light come from

cupoosta [38]

Answer:

It comes from the chemical reaction of The wax and the gas because the wax is made up of carbon-based chemicals so the gas and carbon make the candle light =)

Explanation:

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3 years ago

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What is the sign of the heat change for an exothermic reaction? for an endothermic reaction?

Julli [10]

A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system. The enthalpies of these reactions are less than zero, and are therefore exothermic reactions

3 0

3 years ago

Calculate how many moles of element Q are in 23.53 g of element Q.

pogonyaev

Answer:

0.579 moles

Explanation:

Moles = 23.53/40.64

= 0.5789 moles

7 0

3 years ago

Help, I need this quick. Someone please.

timofeeve [1]

Answer:

the answer to that question is d

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3 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago