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2 years ago

Using your knowledge of symbols, determine which compound listed must contain both copper and oxygen.

Chemistry

1 answer:

Colt1911 [192]2 years ago

4 0

Answer:

Malachite

Explanation:

Malachite is the only listed compound that must contain copper and oxygen.

Copper and oxygen are both elements found on the periodic table. They have the following symbols;

Copper = Cu

Oxygen = O

From the given choices, only option 1 has the symbol Cu and O.

So only malachite contains both copper and oxygen.

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2 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0

2 years ago

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate

nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.

3 0

3 years ago