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3 years ago

11

given the following reaction: CuO (s) + H2 (g) to Cu(s) + H2O (g). If 250 L of hydrogen gas are used to reduce copper (II) oxide

at STP, what mass of copper is to be expected?

1 answer:

QveST [7]3 years ago

4 0

24g is needed no explanation

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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

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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

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3 years ago

If you have a heartburn, how you could you safely help the situation

Ket [755]

The answer is:

:Avoid lying down after eating

That should help

7 0

4 years ago

A 7th grade science class placed three large grade A eggs into beakers and recorded the mass. They then filled one beaker with v

lisov135 [29]

Independent- The vinegar,corn syrup, distilled water
Dependent- The mass of the egg
Constant- the egg and the beaker

3 0

3 years ago

G you have 57.0 ml of a 0.400 m stock solution that must be diluted to 0.100 m. assuming the volumes are additive, how much wate

Amanda [17]

Answer:
added water = 171 ml

Explanation:
Assuming volumes are additive, the rule that we will use to solve this question is:
M1V1 = M2V2
where:
M1 is the initial concentration = 0.4 m
V1 is the initial volume = 57 ml
M2 is the final concentration = 0.1 m
V2 is the final volume that we want to calculate
Substitute with the given in the above equation to get V2 as follows:
M1V1 = M2V2
(0.4)(57) = (0.1)V2
22.8 = 0.1V2
V2 = 228 ml

Now, the final volume is equal to the initial volume plus the amount of added water. So, to get the amount of added water, we will subtract the initial volume from the final volume as follows:
V2 = V1 + added water
228 = 57 + added water
added water = 228 - 57 = 171 ml

Hope this helps :)

3 0

3 years ago