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Mashcka [7]
3 years ago
12

Mention any two uses of oxygen gas.​

Chemistry
1 answer:
Elodia [21]3 years ago
3 0

Answer:

steel, plastics

<h3>Explanation:</h3>

Hope it helps!

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21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4
bija089 [108]
The balanced equation for the above reaction is 
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2  mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
5 0
3 years ago
What type of star has an absolute brightness of 5 and a surface temperature around 3,000 °C?
pav-90 [236]
From the graph. the answer is
A. supergiants
3 0
3 years ago
Read 2 more answers
What is the resultant pressure if 1.7 mol of ideal gas at 273 K and 2.79 atm in a closed container of constant volume is heated
dedylja [7]

Answer: The resultant pressure is 3.22 atm

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T     (At constant volume and number of moles)

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure of gas  = 2.79 atm

P_2 = final pressure of gas  = ?

T_1 = initial temperature of gas  = 273K

T_2 = final temperature of gas = 315 K

\frac{2.79}{273}=\frac{P_2}{315}

P_2=3.22atm

Thus the resultant pressure is 3.22 atm

6 0
2 years ago
By studying fossils, scientists have learned that A. Both animals and plants have changed over time. B. Plants have changed over
Nastasia [14]

Answer:A. Both animals and plants have changed over time.

Explanation:

Because plants and animals go through adaptations.

4 0
3 years ago
Read 2 more answers
If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it
Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

NO_2\rightarrow NO+\frac{1}{2}O_2

The reaction is second order for NO_2 with a rate constant of 0.543M^{-1}s^{-1} at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.543M^{-1}s^{-1}

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

[A]_o = Initial concentration = 0.260 M

Putting values in above equation, we get:

0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s

Hence, the time taken is 5.19 seconds

6 0
3 years ago
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