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2 years ago

Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.

2 answers:

8 0

The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.

jonny [76]2 years ago

7 0

Answer:

248.9 uma

Explanation:

We calculate the average atomic mass using the following formula

$m= \frac{(m_1.\%_1)+(m_2.\%_2)}{100} \\$

This formula depends on how many isotopes the atom has.

If the atom had three isotopes another term would be added to the formula

$m1= 248.7 uma\\\%_1= 75.47\%\\m_2= 249.4 uma\\\%_2= 24.53\%$

$m= \frac{(248.7)(75.47)+(249.4)(24.53)}{100}\\ m=248.9 uma$

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Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a

kirill115 [55]

M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)

5 0

3 years ago

What is an animals life in a desert?

rodikova [14]

An animals life in a desert is to survive, depending on the type of desert though, but I assume the most common desert, the hot/dry desert. Most animals are nocturnal, because it becomes cooler at night, and live underground during the day.

8 0

3 years ago

All of the following are true statements about catalysts except: Select one: a. A catalyst will speed the rate-determining step

liberstina [14]

<span>d. A catalyst will be used up in the reaction is wrong,
amount of catalyst should not change</span>

6 0

3 years ago

Compare the mass of the Mg ribbon with the mass of the magnesium oxide. Notice that the the mass of the magnesium oxide is great

Artist 52 [7]

Answer:

Magnesium oxide is a binary compound of magnesium and oxygen while magnesium ribbon consists only of magnesium atoms.

Explanation:

The burning of magnesium in oxygen is a chemical change. It produces magnesium oxide having greater mass than magnesium ribbon. The greater mass results from the fact that the chemical reaction has added another element to the sample- oxygen. The mass of magnesium ribbon is the mass of magnesium atoms alone but in magnesium oxide, we consider the masses of magnesium and oxygen atoms making magnesium oxide heavier than magnesium ribbon.

4 0

3 years ago

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and

IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x. So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0

3 years ago