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2 years ago

What do you put in the calculator to get the molar mass of Mg3P2

Chemistry

1 answer:

olga55 [171]2 years ago

6 0

Input the atomic masses of Mg and P to give 134.84g/mol

Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

P = 30.97g/mole

Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

learn more:

Molar mass brainly.com/question/2861244

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The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab

mixas84 [53]

Answer:

See the answers below

Explanation:

1) 100. mL of solution containing 19.5 g of NaCl (3.3M)

2) 100. mL of 3.00 M NaCl solution (3 M)

3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)

4) Number 1 and 5 have the same concentration (1.5M)

MW of NaCl = 23 + 36 = 59 g

For number 3

59 g ------------------- 1 mol

19,5 g ----------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For number 4,

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For number 5

Molarity (M) = 0.450/0.3 = 1.5 M

4 0

3 years ago

Read 2 more answers

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$