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2 years ago

How many moles of atoms are in each elemental sample?

Chemistry

1 answer:

Vlada [557]2 years ago

8 0

Answer:

Moles = mass(m) / Molar mass (M)

A. 1.24g of Zn

m = 1.24g

M = 65.4 g/mol

moles of Zn = 1.24g/ 65.4g/mol

= 0.01896mol

B. 28.1 g Ar

m = 28.1g

M(Ar) = 39.9g/mol

moles of Ar = 28.1g/39.9g/mol

= 0.7043 mol

C. 84.0 g Ta

m = 84.0g

M(Ta) = 181g/mol

moles of Ta = 84.0g/181g/mol

= 0.4641 mol

D. 2.29x10^-2 g Li

m = 2.29x10^-2 g

M(Li) = 6.9g/mol

moles of Li = 2.29x10^-2 g / 6.9g/mol

= 0.0033 mol

Hope this helps

You might be interested in

How many sodium atoms will create an ionic bond with sulfur?

Galina-37 [17]

Hi,

Two sodium atoms are needed to create an ionic bond with sulfur.

5 0

3 years ago

A runner completes a 5-mile run.how many yards did she run

slega [8]

The runner ran a total of 8800 yards.

5 0

3 years ago

1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.

svet-max [94.6K]

Answer:

(a) Theoretical Yield = 10.50 g

(b) %age yield = 27.33 %

Explanation:

Answer-Part-(a)

The balance chemical equation for the synthesis of Ammonia is as follow;

N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

Moles = Mass / M/Mass

Moles = 9.75 g / 28.01 g/mol

Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

Moles = Mass / M/Mass

Moles = 1.86 g / 2.01 g/mol

Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

1 mole of N₂ reacts with = 3 moles of H₂

So,

0.348 moles of N₂ will react with = X moles of H₂

Solving for X,

X = 3 mol × 0.348 mol / 1 mol

X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

3 mole of H₂ produces = 2 moles of NH₃

So,

0.925 moles of H₂ will produce = X moles of NH₃

Solving for X,

X = 2 mol × 0.925 mol / 3 mol

X = 0.616 mol of NH₃

Step 5: Calculating theoretical yield of Ammonia as,

Theoretical Yield = Moles × M.Mass

Theoretical Yield = 0.616 mol × 17.03 g/mol

Theoretical Yield = 10.50 g

Answer-Part-(b)

%age yield = Actual Yield / Theoretical Yield × 100

%age yield = 2.87 g / 10.50 g × 100

%age yield = 27.33 %

4 0

3 years ago

Andi performs the calculation that is shown below.

Sergio039 [100]

Answer: -

A) 3.59

Explanation: -

The number of significant figures in 2.06 = 3

The number of significant figures in 1.743 = 4

The number of significant figures in 1.00 = 3

During multiplication of significant figures, the number of significant figures in the answer would be the smallest value that is 3 in this case.

2.06 x 1.743 x 1.00 = 3.5908.

The answer to 3 significant figures is

2.06 x 1.743 x 1.00 = 3.59

Thus, Andi when performs the calculation that is shown below.

(2.06)(1.743)(1.00) the answer be reported using the correct number of significant figures as A) 3.59

8 0

3 years ago

Read 2 more answers

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

2 years ago