State the coefficient required to correctly balance the following chemical equation:

What is the molar mass for a solute if 32.38 g of the solute dissolved in 296 g of water

Pie

Answer:

this is math

Explanation:

7 0

3 years ago

What is the Volume of the salt solution?

hammer [34]

<h3>Answer:</h3>

61.3 mL

<h3>Explanation:</h3>

From the data given;

Mass of volumetric flask is 109.1 g
Mass of salt is 5.01 g
Mass of Volumetric flask and salt is 113.5 g
Mass of volumetric flask and salt solution is 170.4 g
Mass of salt solution is 61.3 g

We are required to calculate the volume of salt solution;

We need to know the relationship between density, mass and volume of a solution.

Density = Mass ÷ Volume

Therefore, given mass and density we can find the volume.

Rearranging the formula;

Volume = Mass ÷ density

Assuming the density of salt solution is 1 g/mL

Then;

Volume = Mass of the salt solution ÷ density of the salt solution

= 61.3 g ÷ 1 g/mL

= 61.3 mL

Therefore, the volume of the salt solution is 61.3 mL

3 0

3 years ago

A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4884 g of a mixture of Al(OH)3 and Mg(OH)2 is

Nitella [24]

Answer:

54.7%

Explanation:

The reaction of neutralization will be:

Al(OH)₃ + Mg(OH)₂ + 5HNO₃ → Al(NO₃)₃ + Mg(NO₃)₂ + 5H₂O

The number of moles of HNO₃ added is the concentration multiplied by the volume (17.12 mL = 0.01712 L)

n = 1*0.01712 = 0.01712 mol

For the stoichiometry:

1 mol of Al(OH)₃ ---------- 5 moles of HNO₃

x ----------------------------------0.01712 mol

By a simple direct three rule:

5x = 0.01712

x = 3.424x10⁻³ mol of Al(OH)₃

For the stoichiometry, the number of moles of Mg(OH)₂ will also be 3.424x10⁻³.

The molar masses are:

Al(OH)₃: 27 g/mol of Al + 3*16 g/mol of O + 3*1 g/mol of H = 78 g/mol

Mg(OH)₂: 24 g/mol of Mg + 2*16 g/mol of O + 2*1 g/mol of H = 58 g/mol

The mass is the molar mass multiplied by the number of moles:

Al(OH)₃: 78*3.424x10⁻³ = 0.2671 g

Mg(OH)₂: 58*3.424x10⁻³ = 0.1986 g

The mass % of Al(OH)₃ in the mixture is its mass divides by the mass of the mixture, multiplied by 100%:

(0.2671/0.4884)x100% = 54.7%

5 0

3 years ago

CO2 and H2O are both compounds. How do you know

Eva8 [605]

Answer:

neutral groups of atoms formed by covalent bonds.

3 0

3 years ago

Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inl

Tasya [4]

Answer:

The exit temperature of the Nitrogen would be 331.4 K.
The area at the exit of the diffuser would be $7*10^{-3} m^2$ .
The rate of entropy production would be 0.

Explanation:

First it is assumed that the diffuser works as a isentropic device. A isentropic device is such that the entropy at the inlet is equal that the entropy T the exit.
It will be used the subscript 1 for the inlet conditions of the nitrogen, and the subscript 2 for the exit conditions of the nitrogen.
It will be called: v the velocity of the nitrogen stream, T the nitrogen temperature, V the volumetric flow of the specific stream, A the area at the inlet or exit of the diffuser and, P the pressure of the nitrogen flow.
It is known that for a fluid flowing, its volumetric flow is obtain as: $V=v*A$ ,
Then for the inlet of the diffuser: $V_1=v_1*A_1=282\frac{m}{s}*4.8*10^{-3}m^2=1.35\frac{m^3}{s}$
For an ideal gas working in an isentropic process, it follows that: $\frac{T_{2} }{T_1}=(\frac{P_2}{P_1})^k$ where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $T_2$ , the temperature of the nitrogen at the exit conditions: $T_2=T_1(\frac{P_2}{P_1})^k$ then, $T_2=300 K (\frac{0.9 bar}{0.656 bar})^{(\frac{1.4-1}{1.4})}=331.4 K$
Also, for an ideal gas working in an isentropic process, it follows that: $\frac{P_2}{P_1}= (\frac{V_1}{V_2})^k$ , where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $V_2$ the volumetric flow at the exit of the diffuser: $V_2=V_1*\frac{1}{\sqrt[k]{\frac{P_2}{P_1}}}=\frac{1.35\frac{m^3}{s}}{\sqrt[1.4]{\frac{0.9bar}{0.656bar} }}=1.080\frac{m^3}{s}$ .
Knowing that $V_2=1.080\frac{m^3}{s}$ , it is possible to calculate the area at the exit of the diffuser, using the relationship presented in step 4, and solving for the required parameter: $A_2=\frac{V_2}{v_2}=\frac{1.08\frac{m^3}{s} }{140\frac{m}{s}}=7.71*10^{-3}m^2$ .
To determine the rate of entropy production in the diffuser, it is required to do a second law balance (entropy balance) in the control volume of the device. This balance is: $S_1+S_{gen}-S_2=\Delta S_{system}$ , where: $S_1$ and $S_2$ are the entropy of the stream entering and leaving the control volume respectively, $S_{gen}$ is the rate of entropy production and, $\Delta S_{system}$ is the change of entropy of the system.
If the diffuser is operating at stable state is assumed then $\Delta S_{system}=0$ . Applying the entropy balance and solving the rate of entropy generation: $S_{gen}=S_2-S_1$ .
Finally, it was assume that the process is isentropic, it is: $S_1=S_2$ , then $S_{gen}=0$ .

6 0

3 years ago