What is this please helpme

How are electric fields different from gravitational fields?

Anni [7]

Yes, using the method of pemdas can reroute the gps

5 0

3 years ago

If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)

Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

$w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%$

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0

3 years ago

A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig

Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

PV = nRT

The given data is as follows.

Pressure, P = 1500 psia, Temperature, T = $104^{o}F$ = 104 + 460 = 564 R

Volume, V = 2.4 cubic ft, R = 10.73 $psia ft^{3}/lb mol R$

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

n = $\frac{mass}{\text{molar mass}}$

m = $n \times W$

= $0.594 \times 16.04$

= 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

We will calculate the density as follows.

d = $\frac{PM}{RT}$

= $\frac{1500 \times 16.04}{10.73 \times 564}$

= 3.975 $lb/ft^{3}$

Now, calculate the specific gravity of the gas as follows.

Specific gravity relative to air = $\frac{\text{density of methane}}{\text{density of air}}$

= $\frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}$

= 51.96

6 0

3 years ago

The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the en

Vesna [10]

Answer:

3.5 atm

Explanation:

As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.

π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.

The difference in osmotic pressure of the solutions is:

Δπ = Δ c RT where c is the difference in molar concentrations.

pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K

= 3.47 atm

3 0

3 years ago

Which is a practical application of boiling-point elevation?

pishuonlain [190]

Boiling-point is the point of a pure liquid matter starts to evaporate and change into gaseous phase. It is where the set of conditions such as the pressure and temperature enough to do so. Boiling-point elevation, on the other hand, is the phenomenon of which the boiling point of a pure liquid matter is elevated because of the dissolved substances. A great example would be the boiling point of a distilled water (pure water) which is lesser than the boiling point of a sea water because of the dissolved salts. A pure water boils at 100°C at atmospheric pressure while a salt water boils at higher temperature than 100°C at the same pressure. Thus, the answer is D.

7 0

3 years ago