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Semenov [28]
3 years ago
7

1. Explain the two processes that keep atmospheric oxygen and carbon dioxide at stable levels?

Chemistry
1 answer:
lesya [120]3 years ago
4 0

Cellular Respiration and Photosynthesis. Photosynthesis is the process of when plants use sunlight to make foods from carbon dioxide and water, to later on make oxygen. Cellular respiration is the process through which cells convert sugars into energy.

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Which of the following is true about the differences between the "Plum Pudding" model and Rutherford's model? There are two poss
Ugo [173]
I think it is the letter A because they both Make sense
7 0
3 years ago
Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(
sesenic [268]
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l),  Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s) 
we are going to take as 
 Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>

</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
4 0
3 years ago
What volume of oxygen at STP is requieres for the complete combustion of 100.50 mL of C2H2
malfutka [58]

The volume of oxygen at STP required would be 252.0 mL.

<h3>Stoichiometic problem</h3>

The equation for the complete combustion of C2H2 is as below:

2C_2H_2 + 5O_2 --- > 4CO_2 + 2H_2O

The mole ratio of C2H2 to O2 is 2:5.

1 mole of a gas at STP is 22.4 L.

At STP, 100.50 mL of C2H2 will be:

                 100.50 x 1/22400 = 0.0045 mole

Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles

0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL

Thus, 252.0 mL of O2 gas will be required at STP.

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

7 0
1 year ago
What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

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3 years ago
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2 years ago
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