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aivan3 [116]
3 years ago
8

3. Nitrogen, when combined with oxygen, can form various oxides. 3.5 g of nitrogen gas is burned in oxygen to form 11.5 g of an

oxide of nitrogen.
a) Determine the mass of oxygen burned. (Hint: The oxide contains ONLY nitrogen and oxygen!)
b)Using the data from the experiment, find the percent of nitrogen by mass in the oxide.
c) Using the data from the experiment, find the percent of oxygen by mass in the oxide.
Chemistry
1 answer:
Pavel [41]3 years ago
5 0

Answer:

A.

Explanation:

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Science help for 6th grade advanced?
tino4ka555 [31]
16287.50 I think? I just googled it though so I’m not sure if it’s correct.
6 0
3 years ago
Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and
alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

                                = 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0
2 years ago
"how many grams of calcium are consumed when 156.8 ml of oxygen"
Akimi4 [234]
Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.

To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,

Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =<em> 0.56 g Ca</em>

3 0
3 years ago
An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

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laila [671]

Answer:

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