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3 years ago

Convert 3 moles of h2o into grams

1 answer:

4 0

To convert 3 mol H2O to grams, just multiply by the molar mass of H2O.

<span>3 mol H2O * 18 g H2O / 1 mol H2O = 54 g H2O

Hope this helped.</span>

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Acid base reaction is called nuetralization reaction why?

Lesechka [4]

Answer:

Explanation:

You have an acid that is acidic or a base that is basic. When you mix the two, they form water (assuming those are bronsted-lowry acids and bases) which is neutral.

3 0

3 years ago

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Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

Propane burns in oxygen to produce carbon dioxide and steam. The unbalanced equation for this reaction is: C3H8 + O2 → CO2 + H2O

NemiM [27]

Here is your answer

C3H8 + O2-----> CO2 + H2O

Balancing C atoms

C3H8 + O2------> 3CO2 + H2O

Balancing H atoms

C3H8 + O2------> 3CO2 + 4H2O

Balancing O atoms

C3H8 + $\bold{5}$ O2-------> $\bold{3}$ CO2 + $\bold{4}$ H2O

Hence the equation is balanced.

HOPE IT IS USEFUL

8 0

3 years ago

Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.

kakasveta [241]

Answer:

The volume will be 89.6875 ml

Explanation:

So to count this we will use a single proportion.

0.0640 mol - 1000 ml

5.74×10−3 mol - x ml

x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml

6 0

3 years ago

(Please answer as quickly as possible for brainliest, thanks :))

Eddi Din [679]

A. actinides do not occur in nature.

4 0

3 years ago

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