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photoshop1234 [79]
2 years ago
9

Will the composition of water molecules vary depending on their source?

Chemistry
1 answer:
Alecsey [184]2 years ago
6 0

Answer:

It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.

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I searched it up but it keeps saying chemical energy, but in the option doesn’t have it...I’m torn between B and C...helppppp!!!
jeyben [28]

Answer: The correct answer is in chemical bonds

Explanation:

When coal is burnt, these components burn and release energy. The energy released is by the chemical reaction between the constituents and oxygen

8 0
2 years ago
How many grams of nitrogen dioxide are required to produce 5.89x10^3 kg of hno3 in excess water
polet [3.4K]
<span>3 NO2 + H2O -------->. 2 HNO3. + NO
3(46g)------------------------> 2 ( 63g) HNO3
? kg-------------------------5.89 x10^3kg HNO3
Mass of NO2. = 5.89x10^3 x 138/ 2(63) = 6.45 x10^3 kg</span>
5 0
3 years ago
CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

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

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
How many grams of sodium are in .500 of a mole
Step2247 [10]

Convert mole to gram by multiplying the molar mass of sodium

0.500mol Na x 22.990g = 11.495g of Na

4 0
3 years ago
Which of these molecules is nonpolar?<br> a.CH3CI<br> b.CO<br> c.O2<br> d.PF3
valentina_108 [34]

Answer:

i think the answer is D

Explanation:

7 0
2 years ago
Read 2 more answers
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