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kirill [66]
3 years ago
9

The table shows the number of atoms of different elements that make up molecules of four compounds. Which compound has the chemi

cal formula C₄H₄N₂O₂
Chemistry
1 answer:
Ann [662]3 years ago
4 0

umm what?

................

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What job can you do without going to college?
Law Incorporation [45]

Retail

entrepreneur

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8 0
3 years ago
They these
makkiz [27]

Explanation:

For number 9. the answer is 668.23255199999

For number 10. the answer is 1029.7634968

For number 11. the answer is 856.7084

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4 0
3 years ago
How many grams of CaOH2 are in this solution? The molar mass of CaOH2 is 58.093 g/mol. (0.0787 mol CaOH2) (58.093 g/mol) = g CaO
Alenkinab [10]

Answer:

mass is 4.57 g

Explanation:

Ca(OH)₂ solution consists of Ca(OH)₂ solute molecules and solvent and we are asked to find the mass of Ca(OH)₂ in the solution

number of moles of Ca(OH)₂ is - 0.0787 mol

molar mass of Ca(OH)₂ is - 58.093 g/mol

we can use the following equation

number of moles = mass of Ca(OH)₂ / molar mass of Ca(OH)₂

rearranging the equation

mass of Ca(OH)₂ = number of moles x molar mass

 mass = 0.0787 mol x 58.093 g/mol = 4.57 g

mass of Ca(OH)₂ is 4.57 g

4 0
3 years ago
What is the volume of 23.4 grams of nitrogen gas at 750 mmHg and 28 degrees C
ratelena [41]

Answer:

V = 20.97 L

Explanation:

Given data:

Mass of nitrogen = 23.4 g

Pressure of gas = 750 mmHg (750/760 = 0.99 atm)

Temperature of gas = 28°C (28+273= 301K)

Volume of gas = ?

Solution:

First of all we will calculate the number  of moles of gas.

Number of moles = mass / molar mass

Number of moles = 23.4 g/ 28 g/mol

Number of moles = 0.84 mol

Now we will calculate the volume:

PV = nRT

V = nRT/P

R = general gas constant = 0.0821 atm.L/mol.K

V = 0.84 mol× 0.0821 atm.L/mol.K× 301 K /  0.99 atm

V = 20.76 L/0.99

V = 20.97 L

6 0
4 years ago
Just a little help with gas laws?
ICE Princess25 [194]

Answer:

You would use the second formula! you can tell because you have an initial volume and a second volume (after blowing it up).

5 0
3 years ago
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