Plz Answer all three questions thx ;)

The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:

Initial volume * initial concentration = final volume * final concentration

0.5 mL * 10M = 5mL * final concentration

1.1 M = final concentration = main solution concentration

Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.

Dilution factor = Main solution concentration/tube 4 concentration

Dilution factor = 1.1/0.0088 = 125

I hope my answer helps you

3 0

3 years ago

Which statements are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain o

Dvinal [7]

Answer:A, C, D, E, F

Explanation:

A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces

B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.

C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.

D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)

E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.

F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)

4 0

3 years ago

A mixture of MgCl2 and NaCl has an initial mass of 0.4015 g. Excess AgNO3(aq) was added and 1.032 g of precipitate, AgCl(s), was

zlopas [31]

Answer:

8.909*10^-4 moles

Explanation:

The mixture contains MgCl $_{2}$ and NaCl with a total mass of 0.4015 g. The mass of precipitate AgCl(aq) is 1.032 g and it has a molar mass of 143.32 g/mol. Therefore, the moles of the precipitate is:

n = 1.032/143.32 = 7.2007*10^-3 moles

Molar mass of NaCl = 58.44 g/mol and the molar mass of MgCl $_{2}$ is 95.21 g/mol. Let the mass (g) of MgCl $_{2}$ in the original mixture be 'x'. Thus:

7.2007*10^-3 = (0.4015-x)/58.44 + 2x/95.21

(7.2007*10^-3)*58.44*95.21 = 95.21(0.4015-x) + 2x(58.44)

40.06505 = 38.227 -95.21x + 116.88x

40.06505 - 38.227 = -95.21x + 116.88x

1.83805 = 21.67x

x = 1.83805/21.67 = 0.0848 g

moles of MgCl $_{2}$ = 0.0848g/95.21 g/mol = 8.909*10^-4 moles

4 0

3 years ago