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elena-14-01-66 [18.8K]
3 years ago
5

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

Chemistry
1 answer:
Pani-rosa [81]3 years ago
5 0

Answer:

666,480 Joules or 669.48 kJ

Explanation:

We are given;

  • Volume of water as 2.0L or 2000 ml

but, density of water is 1 g/ml

  • Therefore, mass of water is 2000 g
  • Initial temperature as 20 °C
  • Final temperature as 99.7° C

Required to determine the heat change

We know that ;

Heat change = Mass × Temperature change × specific heat

In this case;

Specific heat of water is 4.2 J/g°C

Temperature change is 79.7 °C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

                      = 669,480 Joules 0r 669.48 kJ

Thus, the heat change involved is 666,480 Joules or 669.48 kJ

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A researcher wants to determine if a unicellular organism he discovered is an autotroph or a heterotroph. He radioactively label
GaryK [48]

Answer:

A

Explanation:

Autotrophs utilize the energy from  sunlight to reduce carbon dioxide to carbohydrates (glucose). The energy from the sunlight is used to split water into H+ and O2- and the H+ used in the reduction process. The labeled carbon in the carbon dioxide will, therefore, be incorporated by the autotrophs in the carbohydrates made in photosynthesis.  

4 0
2 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
Insulators prevent the loss of heat. A company is designing a mug that will keep coffee warm for a long time. Which one of these
Mumz [18]
Rubber, because it doesn’t conduct heat and it is a good insulator
3 0
2 years ago
When a redox reaction that takes place in an acidic solution involves an oxygen imbalance, oxygen should be balanced by adding _
olya-2409 [2.1K]

Balancing redox reactions:

Oxygen should be balanced by adding H_{2}O  as needed, while hydrogen should be balanced by adding H^{+}.

What is a redox reaction?

Redox reactions, also known as oxidation-reduction reactions, involve the simultaneous oxidation and reduction of two different reactants.

The Half-Equation Method is one technique used to balance redox processes. The equation is divided into two half-equations using this technique: one for oxidation and one for reduction.

By changing the coefficients and adding H_{2}O, H^{+}, and e^{-} in that order, each reaction is brought into equilibrium:

  1. By putting the right number of water (H_{2}O) molecules on the other side of the equation, the oxygen atoms are brought into balance.
  2. By adding H^{+} ions to the opposing side of the equation, one can balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom).
  3. Total the fees for each side. Add enough electrons (e^{-}) to the more positive side to make them equal. (As a general rule, e^{-} and  H^{+}are nearly always on the same side.)
  4. The e^{-} on either side must be made equal; if not, they must be multiplied by the lowest common multiple (LCM) in order to make them equal.
  5. One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
  6. Now that the equation has been verified, it can be balanced.

Learn more about redox reaction here,

brainly.com/question/20068208

#SPJ4

7 0
2 years ago
Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2,
sdas [7]

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O  = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54



4 0
3 years ago
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