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2 years ago

. A compound is composed of 2 atoms of Copper, 1 atom of Sulfur and 4 atoms of Oxygen. What

Chemistry

1 answer:

Dennis_Churaev [7]2 years ago

3 0

The chemical formula for this compound : Cu₂SO₄-Copper(I) sulfate

<h3>Further explanation</h3>

Given

2 atoms of Copper, 1 atom of Sulfur and 4 atoms of Oxygen.

Required

The chemical formula

Solution

The number of atoms forming a compound is usually indicated from the subscript in the formula

From the problem it can be seen that the compound has:

2 atoms of Cu, 1 atom of S, and 4 atoms of O

So the formula:

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

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

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

7 0

3 years ago

Item 3

Artist 52 [7]

Answer:

Speed is measured over time.

Explanation:

7 0

2 years ago

Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?

konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

6 0

2 years ago

51.7ml at 27 Celsius and 90kpa to stp

never [62]

STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K* 90^10^3Pa/ 300.15K * 1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml

6 0

3 years ago