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erastova [34]
3 years ago
5

Calcite is mineral that is found in limestone and marble. A model of the extended structure of calcite is

Chemistry
1 answer:
icang [17]3 years ago
7 0

Answer:

B. It is formed by a repeating pattern of a single type of atom

Explanation:

i think thats answer may i get brainlest?

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Al + AgNO3 -> Al(NO3)3 + Ag. As a balanced equation​
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Al + 4AgNO3 >>Al(NO3)3+ 3Ag

Explanation:

the number of moles of No3 of the products is 3 therefore we have to balance the reactants by adding 3 before the "AgNO3" which also leades us to adding 3 mols to Ag on the products side

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A gas mixture contains oxygen, nitrogen, and carbon dioxide. It has a pressure of 250 mmHg. If the pressure of oxygen is 50 mmHg
kondor19780726 [428]

Answer:

P_{CO_2}=25mmHg

Explanation:

Hello there!

In this case, according to the Dalton's law, which explains that the total pressure of a gaseous system equals the sum of the partial pressures of the gases composing, for the gaseous mixture composed by oxygen, nitrogen and carbon dioxide it would be possible to write:

P_{tot}=P_{N_2}+P_{O_2}+P_{CO_2}

Now, given the pressure of the system and those of oxygen and nitrogen, we calculate that of carbon dioxide as shown below:

P_{CO_2}=P_{tot}-P_{N_2}-P_{O_2}\\\\P_{CO_2}=250mmHg-50mmHg-175mmHg\\\\P_{CO_2}=25mmHg

Best regards!

4 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
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