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ad-work [718]
3 years ago
10

Which of the following is the correctly balanced chemical equation for the reaction Ca(OH)2 and HNO3

Chemistry
2 answers:
avanturin [10]3 years ago
6 0

Answer:

4th one

Explanation:

took it on edge got a 100

Hope this helped :)

denis-greek [22]3 years ago
4 0
 The  correctly  balanced  chemical equation   for the reaction of  Ca(OH)2 and  HNo3  is as below

Ca(OH)2  + 2HNO3 → Ca(NO3)2   +2H20

that is   1 mole of  Ca(OH)2  reacted  with  2 moles of HNO3  to form  1 mole  of Ca(NO3)2   plus 2 moles of H2O
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When Earth’s axis points away from the sun, the _________ Hemisphere has the longer days of summer.
Llana [10]

Answer: northern hemisphere

Explanation: I looked it up. Plus I took a test with this question and when the teacher went over the answer i got it right.

3 0
2 years ago
1. What does it mean for a substance to be soluble?​
Anni [7]

Answer:

Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. ... Certain substances are soluble in all proportions with a given solvent, such as ethanol in water.

Explanation:

3 0
3 years ago
Read 2 more answers
An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and CO2 f
ladessa [460]

Answer: Mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

For iron:

Given mass of iron = 930 kg = 930000 g  (1kg=1000g)

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

\text{Moles of iron}=\frac{930000g}{56g/mol}=16607mol

The chemical equation for the  production of iron  follows:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

By Stoichiometry of the reaction:

2 moles of iron are  produced by =  1 mole of Fe_2O_3

So, 16607 moles of iron will be produced by = \frac{1}{2}\times 16607=8303moles of Fe_2O_3

Now, calculating the mass of Fe_2O_3 from equation 1, we get:

Mass of Fe_2O_3 = moles\times {\text {molar mass}}=8303\times 160=1328480g=1328kg

Thus mass of Fe_2O_3 required to form 930 kg of iron is 1328 kg

6 0
3 years ago
Morphine has the formula c17h19no3. it is a base and accepts one proton per molecule. it is isolated from opium. a 0.685 −g samp
kotykmax [81]
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole 
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34 
                             = 0.059 g
percent morphine = \frac{mass of morphine}{mass of opium} x 100
                             = \frac{0.059}{0.685} x 100 = 8.6 %   
7 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
2 years ago
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