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3 years ago

7. Which 2 structures, in the digestive system, reabsorb water into the body? _ and __

Chemistry

1 answer:

kykrilka [37]3 years ago

5 0

7.Large intestine and small intestine

8.Stomach

9. Capillaies

22.false
4. idk

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What is the density of salt water?

mel-nik [20]

The density of salt water is 1.025 kg/liters .

6 0

3 years ago

Calculate the pH during the titration of 20.00 mL of 0.1000 M HNO2(aq) with 0.1000 M KOH(aq) after 13.27 mL of the base have bee

Slav-nsk [51]

Answer:

pH = 2.462.

Explanation:

Hello there!

In this case, according to the reaction between nitrous acid and potassium hydroxide:

$HNO_2+KOH\rightarrow KNO_2+H_2O$

It is possible to compute the moles of each reactant given their concentrations and volumes:

$n_{HNO_2}=0.02000L*0.1000mol/L=2.000x10^{-3}mol\\\\n_{KOH}=0.1000mol/L*0.01327L=1.327x10^{-3}mol$

Thus, the resulting moles of nitrous acid after the reaction are:

$n_{HNO_2}=2.000x10^{-3}mol-1.327x10^{-3}mol=6.73x10^{-4}mol$

So the resulting concentration considering the final volume (20.00mL+13.27mL) is:

$[HNO_2]=\frac{6.73x10^{-4}mol}{0.01327L+0.02000L} =0.02023M$

In such a way, we can write the ionization of this weak acid to obtain:

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

So we can set up its equilibrium expression to obtain x as the concentration of H3O+:

$Ka=\frac{[NO_2^-][H_3O^+]}{[HNO_2]}\\\\7.1x10^{-4}=\frac{x^2}{0.02023M-x}$

Next, by solving for the two roots of x, we get:

$x_1=-0.004161M\\\\x_2=0.003451M$

Whereas the correct value is 0.003451 M. Finally, we compute the resulting pH:

$pH=-log(0.003451)\\\\pH=2.462$

Best regards!

4 0

3 years ago

Need help !!!!! ASAP

Korolek [52]

<h2>Hello!</h2>

The answer is:

The new volume is 2.84 L.

$V_{2}=2.84L$

To solve the problem, we need to remember what STP means. STP means that the gas is at standard temperature and pressure, or 273.15 K (0°C) and 1 atm.

Also, we need to use the Combined Gas Law, since the temperature, the pressure and the volume are being changed.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using , Gay-Lussac's Law, Charles's Law, and Boyle's Law.

The law is defined by the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

So, we are given the following information:

$V_{1}=2L\\P_{1}=1atm\\T_{1}=0\°C=273.15K\\P_{2}=80kPa=0.8atm\\T_{2}=37\°C=37+273=310K$

Then, isolating the new volume, and substituting, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\V_{2}=\frac{P_{1}V{1}}{T_{1}}*\frac{T_{2}}{P_{2}}\\\\V_{2}=\frac{1atm*2L}{273.15K}*\frac{310K}{0.8kPa}=2.84L$