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sladkih [1.3K]
3 years ago
14

A copper atom has a mass of 1 06 times 10^-22 g and a penny has a mass of 2.5 g. Use this information to answer the questions be

low. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of copper atoms? g How many moles of copper atoms have a mass equal to the mass of a penny?
Chemistry
1 answer:
amm18123 years ago
8 0

Answer:

Mass of 1 mole of copper is 63.83 g.

0.03916 moles of copper atoms have a mass equal to the 2.5 grams of copper penny.

Explanation:

Mass of 1 copper atom,m = 1.06\times 10^{-22} g

1 mole = 6.022\times 10^{23} atoms

Mass of 1 mole of copper :

= 6.022\times 10^{23} atoms\times 1.06\times 10^{-22} g=63.83 g

Mass of 1 mole of copper = 63.83 g

Mass of copper penny = 2.5 g

Atomic mass of copper = 63.83 g/mol

Moles of copper in 2.5 g of copper penny:

\frac{2.5 g}{63.83 g/mol}=0.03916 mol

0.03916 moles of copper atoms have a mass equal to the 2.5 grams of copper penny.

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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

Answer: A. K(s)+O_2(g)\rightarrow KO_2(s) : decreases

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g) : decreases

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g): no change

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

In reaction CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g), as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g), as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0
3 years ago
The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas
Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm

Density is equal to mass per unit volume.

d=\dfrac{m}{V}

m is mass of the cylinder

V is the volume of the cylinder

V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3

So,

m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g

So, the mass of the cylinder is 1.51 grams.

5 0
3 years ago
The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
2 years ago
Use sedimentary in a sentence
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Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:
kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g =  0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3  = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂   I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0
3 years ago
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