I cannot figure out the answers to number 1 or 2

How many atoms in the pictured molecule can form hydrogen bonds with water molecules anatomy?

garri49 [273]

55..............................

6 0

2 years ago

Dr. Winters decides not to tell the people in his experiment about the potential risks. He

anzhelika [568]

Answer: sorry I cant help you I need the same answer

Explanation:

5 0

3 years ago

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0

3 years ago

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago

Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to

hram777 [196]

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$

In that compartment $[M^{n+}]=[m^{2+}]=8.279 \ M$

The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

= 52.82 m mol

= 0.05280 mol

$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

= 2 x 0.05282

= 0.1056 mole

3 0

3 years ago