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goldenfox [79]
3 years ago
7

What is the systematic name of Mg(NO3)2?A) Manganese nitrite.B) Magnesium nitrite.C) Magnesium nitrate.D) Manganese nitrate.

Chemistry
1 answer:
olganol [36]3 years ago
6 0
Mg is magnesium. NO3 is nitrate. This gives you magnesium nitrate as an answer.
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A 13.00 g sample of citric acid reacts with an excess of baking soda as shown in the equation.
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Answer:

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4 0
1 year ago
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What is the effect on the equilibrium when sodium formate is added to a solution of formic acid? hcooh( aq) + h ( aq) right arro
artcher [175]

There is no effect on the equilibrium when sodium formate is added to a solution of formic acid for hcooh( aq) + h ( aq) right arrow choo –( aq).

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4 0
1 year ago
The radius of a xenon atom is 1.3×10−8cm. a 100-ml flask is filled with xe at a pressure of 1.2 atm and a temperature of 281 k .
vladimir2022 [97]
Radius of Xenon = 1.3Ă—10â’8 cm 
Volume = 100 ml = 0.1 L 
Pressure P = 1.2 atm = 121.59 Kpa 
Temperature = 281 K 
R = Gas Constant = 8.31 J mol^-1 K^-1 
Now find the number of atoms 
PV = nRT => n = PV / RT 
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052 
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.  
n = number of atoms= 0.0052 
N = number of particles 
 Avogadro constant A = 6.02 x 10^23 
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20 
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3 
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24 
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
 Fraction of volume will be = 0.00288 / 0.1 = 0.0288
3 0
3 years ago
Which compound exhibits both cis trans and optical isomerism? A: CH3CH=CHCH2CH3B: CH3CHBrCH=CH2C: CH3CBr=CBrCH3D CH3CH2CHBrCH=CH
irinina [24]
CH3CHBrCH=CBrCH3D
I think
4 0
3 years ago
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
3 years ago
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