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3 years ago

Suppose that 10.1 mL of HNO3 is neutralized

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

6 0

Answer:

0.0251 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Next, the data obtained from the question. This include the following:

Volume of acid (Va) = 10.1mL

Molarity of acid (Ma) =..?

Volume of base (Vb) = 61.9 mL

Molarity of base (Mb) = 0.0041 M

Next, we shall determine the molarity of the acid, HNO3 as follow:

MaVa/MbVb = nA/nB

Ma x 10.1 / 0.0041 x 61.9 = 1

Cross multiply to express in linear form

Ma x 10.1 = 0.0041 x 61.9

Divide both side by 10.1

Ma = (0.0041 x 61.9) /10.1

Ma = 0.0251 M

Therefore, the molarity of HNO3 is 0.0251 M

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How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0

3 years ago

In what industry do fertilizers and pesticides wash off and contaminate water supplies?Construction Oil Transportation Agricultu

KIM [24]

Answer:

The answer is agriculture.

Explanation:

3 0

3 years ago

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 2 atm
T1= 50 C= 323 K (being 0 C= 273 K)
P2= 3.2 atm
T2= ?

Replacing:

$\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}$

Solving:

$T2*\frac{2 atm}{323 K} =3.2 atm$

$T2=3.2 atm*\frac{323 K}{2 atm}$

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

5 0

3 years ago

I don't know which the correct answer it is ???

Virty [35]

When I answered i got D

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3 years ago

The temparature of boiling water cannot be measured by _______ temparature

katrin [286]

Answer:

Temperature of boiling water cannot be measured by a Clinical thermometer The reason behind this is that the range of clinical thermometer varies between only 35° C to 42° C.

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2 years ago

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