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2 years ago

Vinegar is a solution of acetic acid in water. If certain vinegar has a concentration of 1.5% v / v.

Chemistry

1 answer:

lina2011 [118]2 years ago

3 0

2 There is about 1000ml in a liter. 1 multiply 1050 by 0.15.

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What is the [H+] of a solution with a pH of 9.40?

LiRa [457]

Answer:

a) 3.98 x 10^-10

Explanation:

Hello,

In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:

$pH=-log([H^+])$

Hence, solving for the concentration of hydronium:

$[H^+]=10^{-pH}=10^{-9.40}\\$

$[H^+]=3.98x10^{-10}M$

Therefore, answer is a) 3.98 x 10^-10

Best regards.

4 0

3 years ago

Write the balanced equation for the burning of nonane, c9h20, in air.

Ivanshal [37]

C9H20 + 14O2 --> 9CO2 + 10H2O

3 0

3 years ago

Blank is the total disappearance of all members of a species.

Vlad [161]

Answer:

Extinction

Explanation:

When a species becomes extinct, every last specimen has died out

3 0

2 years ago

Read 2 more answers

Solid potassium hydroxide koh decomposes into gaseous water and solid potassium oxide . write a balanced chemical equation for t

galina1969 [7]

<span>The chemical formula is pretty straightforward. 2KOH reacts to produce H2O and K2O. This is the balanced chemical reaction between: Solid potassium hydroxide koh decomposing into gaseous water and solid potassium.</span>

4 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

2 years ago