Which phrase describes the molecular structure and properties of two solid forms of carbon, diamond and graphite?(1) the same mo

Answer:

78.96 g of NaC2H3O2

Explanation:

The following information is provided:

• The solution's volume is 350 mL
• The solution's molarity is 2.75 M
• The molar mass of NaC2H3O2 is 82.04 g/mol

We need to find the mass of the solute:

First, we calculate the number of moles:

Moles = Molarity × Volume

Thus;

Moles of solute = 2.75 M × 0.350 L

                        = 0.9625 moles

Next, we find the mass:

Mass = Moles × Molar mass

        = 0.9625 moles × 82.04 g/mol

        = 78.9635 g

      = 78.96 g

Therefore, the amount of NaC2H3O2 required is 78.96 g

Response:

of can be found in 39.5 grams of .

Clarification:

Atomic weights: P= 31, F= 19,

The molar mass equals 1 atomic weight of P + 5 atomic weights of

 F= 31+5 × 19 = 31+95

=126 g/mole

The number of moles in 39.5 gm of

equals

 =

 =0.3134 moles

1 mole of any substance encompasses

0.3131 moles comprises 0.3134

Thus, of can be found in 39.5 grams of .

The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

10 mL of this solution is further diluted to 250 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

5 mL of this solution is diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 500 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.