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3 years ago

You don't need to answer all of it. The main part I need is the bottom left and the top right boxes answered. Thanks.

Chemistry

2 answers:

ahrayia [7]3 years ago

5 0

The green house effect is the trapping of the sun's heat in a planet's lower atmosphere. However due to the larger transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planet's surface.

denis-greek [22]3 years ago

3 0

The questions are a little hard to read

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12.2 g of N₂ gas at 1132 torr and 91°C has a volume of

Bingel [31]

Answer:

8.73 L

Explanation:

First, you need to convert grams to moles using the molar mass.

Molar Mass (N₂): 2(14.009 g/mol)

Molar Mass (N₂): 28.018 g/mol

12.2 grams N₂ 1 mole
---------------------- x ------------------------ = 0.435 moles N₂
28.018 grams

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (torr)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 torr*L/mol*K)

-----> T = temperature (K)

After converting the temperature from Celsius to Kelvin, you can plug the given values into the equation.

P = 1132 torr R = 62.36 torr*L/mol*K

V = ? L T = 91 °C + 273.15 = 364.15 K

n = 0.435 moles

PV = nRT

(1132 torr)V = (0.435 moles)(62.36 torr*L/mol*K)(364.15 K)

(1132 torr)V = 9888.015

V = 8.73 L

6 0

2 years ago

What is the element of 1s^22s^2p^63s^23p^64s^23d^104p^65s^24d^105p^1

77julia77 [94]

The answer is Indium

7 0

3 years ago

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO(g)

AVprozaik [17]

Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO(g) + NO₂(g) → N₂O₃(g) ΔH o rxn = −39.8 kJ

(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH o rxn = −112.5 kJ

(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

If the chemical equation is inverted, the symbol of ΔH is also reversed.
If the coefficients are multiplied, multiply ΔH by the same factor.
If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) you must do the following:

Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

4 NO₂(g) → 2 N₂O₄(g) ΔH o rxn = −114.4 kJ

Reverse equations (1) and (2)

(1) N₂O₃(g) → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ

(2) N₂O₅(g) → NO(g) + NO₂(g) + O₂(g) ΔH o rxn = 112.5 kJ

Equations (4) and (5) are maintained as stated.

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

8 0

3 years ago

A solution with a volume of 1.00 l is 0.450 m in ch 3 cooh(aq) and 0.550 m in ch 3 coona(aq). what will the ph be after 0.0800 m

Semenov [28]

Before addition of HCl,

conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M

After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl

Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)

Now, conc. of H+ ions = ka $\frac{[acid]}{[conjugated base]}$
where ka = dissociation constant for acid = 10^-5 for Ch3COOH

∴ conc. of H+ ions = $\frac{0.53}{0.47}$
= 1.1277 x 10^-5

Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94

4 0

3 years ago

Which pH is the most acidic

sesenic [268]

C. 0.5

This is the answer.

(assuming that the D option is 1, if it is 0.1 then the answer is 0.1)

6 0

3 years ago

Read 2 more answers