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blondinia [14]
3 years ago
12

An element occurs naturally as a gas. It is also characterized by a very

Chemistry
1 answer:
Delvig [45]3 years ago
8 0
I believe it would be c

Please brainiest
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The nuclear binding energy of one nitrogen-14 atom is (BLANK) x 10A J. Round to 3
d1i1m1o1n [39]

mass defect = mass of constituents - mass of atom

N has 7p and 9n

proton mass ~ 1.00728 amu

neutron mass ~ 1.00866 amu

electron mass ~ 0.000549 amu

Nitrogen mass ~ 14.003074 amu

mass defect = (7*1.00728)-(7*1.00866)-(7*0.000549) - 14.003074

= 0.11235amu

convert to energy, the binding energy = 1.68x10^-11 J


5 0
3 years ago
A type of physical that occurs between two electrically charged particles?
White raven [17]
Electromagnetism

Theoretically it is a branch of physics that contains two sources. Either electrically charged particles or the behavior between Neutrons and protons etc.
7 0
2 years ago
For the reversible, one-step reaction A+B<---->C+D, the forward rate constant is 52.4 /mol*h and the rate constant for the
Fantom [35]

Answer:

—-—1———————-s

7 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0
2 years ago
How many moles are in 2.5 g of N2?<br> 0.089 moles<br> KD 0.18 moles<br> 1.3 moles<br> 11 moles
NARA [144]

Answer:

0.089

Explanation:

2.5/ 14= .178 then 0.178/ 2 again and you get 0.089

6 0
2 years ago
Read 2 more answers
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