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3 years ago

Calcium carbonate+nitic acid

Chemistry

1 answer:

miss Akunina [59]3 years ago

8 0

Answer:

What does calcium carbonate and nitric acid make?

Explanation:

Calcium carbonate reacts with nitric acid to produce carbon dioxide.

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3 H2S + 2 HNO3 +35 + 2 NO + 4 H2O

Anettt [7]

Element Oxidized: S (S^(2-) --> S⁰)

Element Reduced: N (N^(5+) --> N(2+))

Oxidizing Agent: HNO3

Reducing Agent: H2S

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2 years ago

2 Points

mr_godi [17]

Answer:

D. Separation by melting point

7 0

3 years ago

Read 2 more answers

If ya could change somethin bout ya previous science classes what would it be?

k0ka [10]

If I could change something in my previous science classes, it would have been the fact that I struggled in the question concerning exponents. It was simple one, but I don't know how I didn't understood that. For example, if there is a question with a calculation "xy²", then I would first multiply x and y, and then calculate the exponent. I did 3 of the questions wrong because of this, it was when I was in 7th grade. I wish I had not done this kind of silly mistake. I an in 10th grade and understand it well now, though.

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2 years ago

What is the density of a sample of an unknown metal has a mass of 35.4 g and a volume of 3.11 cm3?

aev [14]

Density = Mass/Volume. Rounded to 3 significant figures, the density is 11.4 g/cm3

7 0

3 years ago

Ethene is converted to ethane by the reaction C2H4 1g2 1 H2 1g2 8888n C2H6 1g2 C2H4 flows into a catalytic reactor at 25.0 atm a

irakobra [83]

Answer:

Percentage Yield = 93.8%

Explanation:

Data:

R = 0.08206 L.atm/mol/K,

Molar mass of C2H6 = 30.08 g/mol

T = 300° c +273K = 573.15 K

C2H4 Conditions,

P = 25 atm

V = 1000 L/min

H2 Condition,

P = 25 atm

V = 1500 L/min

Solution:

First we will calculate the moles of ethane produced in one minute,

nC2H6 = (15/30.08) ( 1000/1)

nC2H6 = 498.67 mol/min

Now we will calculate the moles of ethane flowing in one minute using the ideal gas equation which is, Pv = nRT

n = pV/R

nC2H4 = ((25* 1000)/(0.08206*573.15))

nC2H4 = 531.54 mol/min

Similarly Calculating the moles of hydrogenflowing in one minute,

nH2 = ((25*1500)/(0.08206)(573.15))

nH2 = 797.32 mol/mim

Both hydrogen and ethene are flowing at 25 atm and 573.15 K but the flow rate of hydrogen is more than that of ethane which results in less amount of available ethene for the reaction. We know that 1 mole of ethene reacts with 1 mole of hydrogen to give 1 mole of ethane

hence ethene is acting as the limiting reagent here

Now calculating the %Yield = (498.67/531.54) * 100

%Yield = 93.82%

4 0

2 years ago