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DaniilM [7]
3 years ago
6

A 250mL sample of air at STP contains approximately 52.5mL of O2. What is the percentage of oxygen in the air

Chemistry
1 answer:
RSB [31]3 years ago
4 0
Yessirrrrrrr wee litttttttt sirrrrr
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Choose the correct answer from the options given.
Pavel [41]

Answer: 1 part is sulfuric acid is a catalyst

2 part is vanadium (V) oxide is a catalyst

Explanation:

4 0
3 years ago
The population of the Earth at the end of 2018 was approximately 7.7 × 10 9 people. Which of the following is the correct way of
stepladder [879]

Answer:

7700000000

Explanation:

so 7.7 x 10 to the power of 9 is in standard form

to put it into a number you just need to times it by 1000000000

however your answer will only have 8 zeros because of the decimal point

5 0
3 years ago
A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of ch
Norma-Jean [14]

Explanation:

The given data is as follows.

   Current (I) = 3.50 amp,        Mass deposited = 100.0 g

  Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

                z \times 52 g = 96500 \times 100 g

                         z = \frac{96500 \times 100 g}{52 g}

                            = 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

                      Q = I × t

           185576.9 C = 3.50 amp \times t  

                      t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.

3 0
3 years ago
Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

4 0
3 years ago
True or false. (Pls help)
grigory [225]
I believe they’re both true.
5 0
2 years ago
Read 2 more answers
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