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3 years ago

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:

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.

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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

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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:

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.
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).
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.)
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.
One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
Now that the equation has been verified, it can be balanced.

Learn more about redox reaction here,

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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

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3 years ago