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zheka24 [161]
3 years ago
5

I need help pls! :[​

Chemistry
1 answer:
baherus [9]3 years ago
6 0

dm me on brainly so i can elp you

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What mass of radon is in 8.17 moles of radon?
Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon  = 8.17moles

Unknown:

Mass of radon  = ?

Solution:

To solve this problem, we use the expression below:

      Number of moles = \frac{mass}{molar mass}  

Molar mass of radon  = 222g/mol

Now insert the parameters and solve;

    Mass of radon  = Number of moles x molar mass

                              = 8.17 x 222

                              = 1813.74g

4 0
3 years ago
Please help thank you (15 points)
dsp73

Answer:

C.

Explanation:

The arrows represent the Earth spinning on its own axis in this picture.

4 0
3 years ago
What is the starting molecule for glycolysis?
meriva
Glucose is the starting molecule for glycolysis.
5 0
3 years ago
How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +
krek1111 [17]

Answer:  13.9 g of H_2O will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

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

\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles

The balanced chemical reaction is:

4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)

According to stoichiometry :

7 moles of O_2 produce =  6 moles of H_2O

Thus 0.900 moles of O_2 will produce =\frac{6}{7}\times 0.900=0.771moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g

Thus 13.9 g of H_2O will be produced from the given mass of oxygen

5 0
3 years ago
Balance the following chemical reactions:
Tcecarenko [31]

1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

2. CH₄ + 2O₂ → CO₂ + 2H₂O

3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃

1)

NaCl+AgNO_3=>NaNO_3+AgC\downarrow

2)

CuSO_4+Cu_2Cl_2\neq>

CuSO_4+Cu_2l_2\neq

4 0
3 years ago
Read 2 more answers
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