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lions [1.4K]
2 years ago
6

Which of the following procedures is used when refining petroleum? Increase the temperature of the oil. Decrease the temperature

of the oil. Increase the pressure of the oil. Decrease the pressure of the oil.
Chemistry
2 answers:
Novosadov [1.4K]2 years ago
6 0
Increase the tempature
arsen [322]2 years ago
4 0
The temperature increases 
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Which statement is true of a reversible reaction at equilibrium?
Vedmedyk [2.9K]

Answer:

D.

The concentration of reactants and the concentration of products are constant.

Explanation:

pls mark as brainliest

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3 years ago
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Structural formula for:<br>5 - methyl hept - 3 - enal​
leonid [27]

Answer:

draw the carbon chain is containing 6 carbon then attach the aldehyde group with edge carbon in chain then put the put double bond at 3 no. carbon

6 0
2 years ago
How many bananas are equal to 7.50 moles of bananas?​
Veseljchak [2.6K]

Answer:

4.52×10^24

Explanation:

N = n × Na

where; N = no. of bananas

n = no. of moles

Na = Avogadro's constant

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N = 7.5 × 6.02×10^23

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2 years ago
Question 1<br>Write the next three numbers in the sequence<br>2, 10, 50, 250,-<br>​
Bess [88]
1250, 6250, and 31250
8 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
2 years ago
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