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3 years ago

What is the molar mass of 12?

Chemistry

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

I assume your talking about carbon when you say 12 so it'd be 12 grams if you are

Explanation:

The molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units.

Hope this helps you some

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A scientist has 400 ml of an 8% solution. How much water needs to be added to reduce this to a 2% solution?

mr_godi [17]

Answer:

Explanation:30

7 0

4 years ago

A susbtance has an empirical formula of CrO2 and a molecular weight of 168 g/mol. determine the molecualr formula for this subst

Sunny_sXe [5.5K]

Answer:

$Cr_{2} O_{4}$

Explanation:

$\frac{molecular weight}{molar mass}$ = $\frac{168}{83.99}$ = 2.00

$CrO_{2}$ x 2 = $Cr_{2} O_{4}$

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2 years ago

If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many

vladimir2022 [97]

Answers:
Answer 1: 10.03 g of siver metal can be formed.
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation

Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
 2 mol Ag x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

 1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.

8 0

3 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

Answer: The mass of sodium acetate that must be added is 30.23 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago

What is the mass of aluminum metal that releases 22.4 l of hydrogen gas at stp from hydrochloric acid? __al(s) + __hcl(aq) → __a

Keith_Richards [23]

The mass of Al metal that releases 2.4 L of hydrogen gas at STP is calculated as below

AT STP 1 mole =22.4 l
what about 22.4l =? moles
(22.4 L x 1 mole) /22.4 L= 2mole

write the reacing reaction

2Al+ 6 HCL = 2AlCl3 + 3H2
by use of mole ratio between Al to H2 which is 2 : 3 the mole Al = 1 x2/3 = 0.667 moles
mass =moles x molar mass
=0.667 mole x27 g/mol= 18 grams

3 0

4 years ago