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2 years ago

If you have 1.5 grams of silver nitrate, how many grams of silver could you produce?

Chemistry

1 answer:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:

0.95g of Ag we can produce

Explanation:

To solve this question we have to convert the mass of silver nitrate (AgNO₃) to moles. The moles of silver nitrate = Moles of Ag. Then, with molar mass of Ag we can know the mass of silver:

Moles AgNO₃ -Molar mass: 169.87g/mol-:

1.5g * (1mol / 169.87g) = 0.0088 moles Ag

Mass Ag -Molar mass: 107.87g/mol-:

0.0088 moles Ag * (107.87g / mol) =

<h3>0.95g of Ag we can produce</h3>

You might be interested in

Empirical formula for compound of 2.17 mol N and 4.35 mol O

balu736 [363]

Answer:

Explanation:

ratio of moles of N and O in molecule =

N / O = 2.17 / 4.35

1/2

empirical formula = NO₂

4 0

2 years ago

Please fill out completely for my understanding. Thank-You!

Archy [21]

The atomic number,atomic mass,valence electrons and the formula for calculating the number of neutrons is also given.The following table has been completed below in the attachment.

Explanation:

Atomic Number:the number of protons /electrons present in the nucleus of an atom is called the atomic number.

Mass Number/Atomic Mass:the number of protons and neutrons present in a nucleus

Example: $C_{6} ^{12}$ ;here;6 implies the atomic number and 12 implies the mass number/atomic mass

No.of neutrons=Mass number-Number of protons

Valence electrons:the number of electrons present in the outer most shell of an atom are called valence electrons

Example:If we consider a carbon atom;as its electronic configuration is2,4 and it belongs to 4th group.Therefore the number of valency electrons in carbon is 4.

8 0

3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$