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JulijaS [17]
3 years ago
10

) What parts of pizzly resemble that of a grizzly bear?

Chemistry
1 answer:
user100 [1]3 years ago
6 0

Answer:

Animal was shot in Nunavut, Canada earlier this month by hunters. It resembles a polar bear but has claws and brown paws of a grizzly bear. Experts claim the 'pizzly' or 'grolar' bear is a hybrid between the two. DNA tests revealed it was actually a blonde haired grizzly bear

Explanation:

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hodyreva [135]
I think it’s C but I’m not sure
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What substances pass through a leaf's stomata?
Lera25 [3.4K]
Oxygen and carbon dioxide
7 0
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C(S)+O2(g)-->CO2(g)
soldi70 [24.7K]

<u>Answer:</u> The correct answer is 1.18 g.

<u>Explanation:</u>

We are given a chemical equation:

C(S)+O2(g)\rightarrow CO_2(g)

We know that at STP conditions:

22.4L of volume is occupied by 1 mole of a gas.

So, 2.21L of carbon dioxide is occupied by = \frac{1}{22.4L}\times 2.21L=0.0986mol of carbon dioxide gas.

By Stoichiometry of the above reaction:

1 mole of carbon dioxide gas is produced by 1 mole of carbon

So, 0.0986 moles of carbon dioxide is produced by = \frac{1}{1}\times 0.0986=0.0986mol of carbon.

Now, to calculate the mass of carbon, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of carbon = 0.0986 mol

Molar mass of carbon = 12 g/mol

Putting values in above equation, we get:

0.0986mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=1.18g

Hence, the correct answer is 1.18 g.

3 0
3 years ago
Read 2 more answers
Calculate the percent yield when 500 grams of carbon dioxide react with an excess of water to produce 640 grams of carbonic acid
Elena-2011 [213]

Answer:

Percent yield =  90.5%

Explanation:

Given data:

Mass of carbon dioxide = 500 g

Mass of water = excess

Actual yield of carbonic acid = 640 g

Percent yield = ?

Solution:

Balanced chemical equation:

CO₂ + H₂O  → H₂CO₃

Number of moles of carbon dioxide

Number of moles  = Mass / molar mass

Number of moles = 500 g/ 44 g/mol

Number of moles = 11.4 mol

Now we will compare the moles of H₂CO₃ with CO₂.

                              CO₂          :              H₂CO₃

                                 1             :                  1

                               11.4           :                11.4

Mass of carbonic acid:

Mass = number of moles × molar mass

Mass = 11.4 mol × 62.03 g/mol

Mass = 707.14 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield =  640 g/ 707.14 g × 100

Percent yield =  90.5%

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The mass of Hydrogen is 2 g/mol
The mass of Helium is 4 g/mol
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