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2 years ago

The molar mass of an element is the mass of one

Chemistry

1 answer:

allochka39001 [22]2 years ago

7 0

Answer:

The molar mass of an element is the mass of one mole of atoms of the element.

Explanation:

The molar mass of an element is its atomic mass, i.e. the mass in grams of one mole of atoms of the element.

Remember 1 mol is approximately 6.022 × 10²³.

So, 1 mol of atoms is 6.022 × 10²³ atoms.

The molar mass is an average: it is the weighted average mass of the natural isotopes of the element, taking into account their relative abundance.

For example, the molar mass or atomic mass of carbon is 12,0107 g/mol, instead of 12.0000, becasue carbon exists in several forms (isotopes), and so the weighted average is not a whole number.

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Find the mass of oxygen in grams produced by the decomposition of 100.0 g of CO2

SSSSS [86.1K]

The balanced chemical equation is :

$2CO_2->2CO+O_2\\\\$

Moles of $CO_2$ ,

$n = \dfrac{100\ g}{44.01\ g/mol}\\\\n=2.27\ mol$

Now, by given chemical equation , we can see 2 mole of $CO_2$ react with 1 mole of $O_2$ .

So , 2.27 mole react with :

$N=\dfrac{2.27}{2}\ mol\\\\N=1.135\ mol$

Mass of oxygen is :

$M = N \times 16\\\\M=1.135\times 16\ g\\\\M =18.16\ g$

Therefore, mass of oxygen in grams produced is 18.16 g.

Hence, this is the required solution.

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2 years ago

If the heat capacity of an object is known, what other information will need to be known to calculate its specific heat capacity

zimovet [89]

Answer:

B. its temperature

Explanation:

4 0

1 year ago

Read 2 more answers

What mass of magnesium would combine with exactly 16.0 grams of oxygen

TEA [102]

1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO

solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16

24.6 g is the answer.

if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.

3 0

3 years ago

What is the hybridization of the carbon atoms in 1,3-butadiene, h2c=ch-ch=ch2?

Natalija [7]

The main formula is as follow is explained in the attached file (please look at the examples)

the 1,3- butadiene is h2c=ch-ch=ch2, so we have

sp² sp² sp² sp²
h2c = ch - ch = ch2

the hybridization of the carbon atoms is sp² : trigonal planar

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3 years ago

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

Explanation:

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

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

2 years ago