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Bess [88]
3 years ago
13

Which quantity is equivalent to 50 kilocalories

Chemistry
1 answer:
Tanya [424]3 years ago
7 0

I believe this problem have these following choices:

5,000 cal

0.05 cal

5 x 10^3 cal

5 x 10 ^4 cal

 

<span>Now 1 kilocalorie  = 1000 calories, therefore:</span>

 

50 kilocalories * (1000 calories / 1 kilocalorie) = 50,000 calories

 

 

Since there are 4 zeroes, so the scientific notation is:

 

5 x 10 ^4 cal         (ANSWER)

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3 years ago
Need help !!!!! ASAP
Julli [10]
<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

V_{2}=1L

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

P_{1}V_{1}=P_{2}V_{2}

So, we are given the information:

V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa

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P_{1}V_{1}=P_{2}V_{2}

V_{2}=\frac{P_{1}V_{1}}{P_{2}}

V_{2}=\frac{50kPa*2L}{100kPa}=1L

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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

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2 years ago
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