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Whitepunk [10]
2 years ago
12

When exothermic reactions occur, ______________.

Chemistry
1 answer:
marusya05 [52]2 years ago
3 0

Answer:

heat energy is released into the surrounding

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What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
kotykmax [81]
<h2><u>Answer:</u></h2>

0.126 Liters

<h2><u>Explanation:</u></h2>

V = mRT / mmP

First, convert the 2.25g of Nitrogen gas into moles. (m in the equation above)

2.25g x 1 mole / 28.0g = 0.08036 moles = m

28.0g = mm

Next, convert the 273 Celsius into Kelvin. (T in the equation above)

273 Celsius + 273.15 = 546.15K = T

R = 0.08206L*atm/mol*K

(Quick Note: The R changes depending on the Pressure Unit so do not use this number every time.)

Now, plug everything into the equation.

V = (0.08036)(0.08206)(546.15)/(28.0)(1.02)

V = 0.126 L

5 0
3 years ago
Give the % composition for each element found in tin (IV) nitride.
drek231 [11]

32.3699% Tin (Sn)

15.2774% Nitrogen (N)

52.3527% Oxygen (O)

4 0
3 years ago
Which of the following is a scientific question you could ask about the game
aniked [119]

Answer:

Answer is option C

Explanation:

If you like my answer than please mark me brainliest thanks

3 0
3 years ago
What general rule can be followed when choosing a type of solvent for a particular solute
ehidna [41]
General 'rule' - "like dissolves like". The solubility of a solute in a solvent (that is, the extent of the mixing of the solute and solvent species) depends on a balance between the natural tendency for the solute and solvent species to mix and the tendency for a system to have the lowest energy possible.

Hope this helps :p
4 0
3 years ago
One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
3 years ago
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