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3 years ago

If the specific heat of water is 1.0 calorie/gram°c, how many calories are required to raise 500 g of water 10.0°c?

1 answer:

Klio2033 [76]3 years ago

7 0

We can use the equation to find out
E = mc△T
E is the total energy (calories) needed, m for mass, c for specific heat capacity and △T for the temperature change.
Substitute the numbers given
E = 500 x 1 x 10
E =5000
Therefore the answer is 5000Cal

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F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?

alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol C4H8) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2

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3 years ago

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3 years ago

A molecule with 14 total electrons and 12 total protons

Snowcat [4.5K]

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Explanation:

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2 years ago

A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55

Xelga [282]

Answer: The volume of NaOH required is 402.9 mL

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For HCl:

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol$

For sulfuric acid:

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

$0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol$

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

$0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL$

Hence, the volume of NaOH required is 402.9 mL

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3 years ago