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3 years ago

In the future, biotechnology could be used to select traits in unborn babies.

1 answer:

3 0

No, i dont think the idea of that is very good. sure, it could possibly help prevent disabilities and such, but it could also be hazardous. shouldnt you love your baby no matter what traits it has??? i dont believe it should be allowed

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An element has an atomic mass number of 16 and an atomic number of 7

Kitty [74]

Answer: Nitrogen

Explanation: Nitrogen has an atomic number of 7 and an atomic mass of about 14 u

7 0

3 years ago

What causes the Brownian motion of particles in a colloid?

LekaFEV [45]

<u>Answer:</u>

The correct answer option is a) collisions between the particles and surrounding molecules.

<u>Explanation:</u>

The collisions between the particles and surrounding molecules causes the Brownian motion of particles in a colloid.

Brownian motion is the irregular movement of the microscopic particles in a fluid which bombard into each other.

It basically is the result of the molecules of a dispersion medium colliding with the dispersed particles of the phase.

6 0

2 years ago

The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of

tresset_1 [31]

Answer:

The total number of moles of gas in the mixture is 0.16939.

1.25550 grams of methane gas and 3.18848 grams of oxygen gas.

Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n = $n_1+n_2$

Using an ideal gas equation :

$PV=nRT$

$P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K$

$P=2.509 n$

Partial pressure of the methane= $p_1=0.175 atm$

Moles of the methane= $n_1$

Partial pressure of the oxygen gas= $p_2=0.250 atm$

Moles of the methane= $n_2$

Mole fraction of the methane= $\chi_1=\frac{n_1}{n_1+n_2}$

Mole fraction of the oxygen gas= $\chi_2=\frac{n_2}{n_1+n_2}$

$p_1=p\times \chi_1$ (Dalton's law)

$0.175 atm=P\times \frac{n_1}{n_1+n_2}$

$0.175 atm=(2.509 n)\times \frac{n_1}{n}$

$n_1=0.06975 mol$

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

$p_2=p\times \chi_2$ (Dalton's law)

$0.250 atm=P\times \frac{n_2}{n_1+n_2}$

$0.250 atm=(2.509 n)\times \frac{n_2}{n}$

$n_2=0.09964 mol$

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n = $n_1+n_2$

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]

7 0

3 years ago

A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express

Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

$wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63$

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

$V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w$

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

$\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3$

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

$Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3$

We can now calculate the molarity as

$Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}$

8 0

3 years ago

What is the percent compostion of anhydrous salt in the hydrate MgBr2 x 6H2O

docker41 [41]

Answer:

Explanation:

5 0

3 years ago