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lions [1.4K]
3 years ago
8

An aqueous solution turns red litmus solution blue. Excess addition of which of the following solution would reverse the change:

Chemistry
1 answer:
MAVERICK [17]3 years ago
6 0
I think the answer is D
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A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod
Olin [163]

Answer:

The answer is 6.25g.

Explanation:

First create your balanced equation. This will give you the stoich ratios needed to answer the question:

2C8H18 + 25O2 → 16CO2 + 18H2O

Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:

7.72 g / 16 g/mol = 0.482 mol

Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:

x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2

x = 0.347 mol H2O

The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:

0.347 mol x 18 g/mol = 6.25g

8 0
3 years ago
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How many hydrophobic tails does a triacyl glycerol molecule have?
liberstina [14]
Three identical fatty acid tails, or three different fatty acid tails (with different lengths or patterns of double bonds).
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2 years ago
PLEASE NO LINKS, GIVING BRAINLIEST, 18 POINTS , THANKS AND 5 STAR!!!!!!!!
Korvikt [17]

Answer:

2nd one In my opinion....

5 0
3 years ago
An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s2? (Formula: F=ma) 0.83 N 1.2 N 11 N 30 N
masya89 [10]
Answer: Option (D) 30N

Detailed Solution:
According to Newton's second law:

F = ma --- (A)

Given:
mass = 5kg
acceleration = 6 m/s^2
F = ?

Plug all the value in equation (A)
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Ans: F = 30N
4 0
3 years ago
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If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised
alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Given ,  

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}

Solving for V₂ , we get:

<u>V₂ = 15.04 mL</u>

3 0
3 years ago
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