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3 years ago

What is the molarity of a solution if 509.6 grams of AgNO3 is dissolved into 12.5 Liters of water?

Chemistry

2 answers:

mihalych1998 [28]3 years ago

5 0

<h3>REFER THE ATTACHMENT ABOVE</h3>

Reptile [31]3 years ago

4 0

Answer:

Molarity is defined as number of moles of substance per litre. So , to find the molarity of this solution , first find the number of moles in 5.9g of AgNO3… 1 mol of AgNO3 has mass= 108+14+16×3= 170g. Therefore Number of moles in 5.9 grams of AgNO3 comes out to be 5.9/170mol in 20ml… ie.

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3 years ago

As Matter Changes States of matter does chemical composition stay the same or not

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4 years ago

Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide accordin

Rudik [331]

Answer:

The limiting reactant is NaOH (option B)

Explanation:

2S(s) + 3O₂(g) + 4NaOH(aq) → 2Na₂SO₄(aq) + 2H₂O(l)

The reaction is ballanced. OK

We need to know how many moles do we have from each compound.

Mass / Molar weight = Mol

Molar weight S = 32 g/m

Molar weight O₂ = 32 g/m

Molar weight NaOH = 40 g/m

Mol S: 2g/ 32g/m = 0.0625 mol

Mol O₂: 3g / 32 g/m = 0.09375 mol

Mol NaOH: 4g/ 40g/m = 0.1 mol

Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.

2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH

0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH

The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.

Let's do the same with O₂

3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH

0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH

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3 years ago

What is physical weathering?

elena-14-01-66 [18.8K]

Physical weathering is caused by physical processes such as changes in temperature, freezing and thawing, and the effects of wind, rain and waves.

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3 years ago

A polymer sample combines five different molecular-weight fractions, each of equal weight. The molecular weights of these fracti

kakasveta [241]

Answer:

Mn = 43,783

Mw = 60,000

Mz = 73,333

narrow distribution = 1.37

Explanation:

The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. This means their Mi is respectively: (The molar weight (Mi) of the fractions)

Fraction 1 : Mi = 20 *10^3

Fraction 2: Mi = 40 *10^3

Fraction 3 : Mi = 60 *10^3

Fraction 4: Mi = 80 *10^3

Fraction 5 : Mi = 100 *10^3

The ΣMi = 300*10^-3

The Wi (mass of the fractions is for all the fractions the same, let's say 1)

So Wi = 1+1+1+1+1 = 5

Since number of moles = mass / Molar mass

The number of moles is respectively: ni = Wi/Mi (x10^5)

Fraction 1 : ni = Wi/Mi = 1/20000 = 5

Fraction 2: ni = 1/40000 = 2.5

Fraction 3 : ni =1/60000 = 1.67

Fraction 4: ni = 1/80000 = 1.25

Fraction 5 : ni= 1/100000 = 1

The Σni = 11.42

Mn = ΣWi/ni = 5/11.42*10^-5 = 43,783

Mw = (ΣWi * Mi)/ΣWi = 300,000 /5 = 60,000

Mz = (ΣWi * Mi²)/ΣWi *Mi = (4*10^8 +16*10^8 +36*10^8 +64*10^8 +100*10^8) /300,000 =73,333

Mz/Mn = narrow distribution =60,000/43,783 = 1.37

7 0

3 years ago

Read 2 more answers