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2 years ago

A warm front is heading toward Chicago. Which type of weather would one expect when it arrives?

Chemistry

2 answers:

nikklg [1K]2 years ago

5 0

B. warm and sunny
but if there is a cold front already in chicago and they collide it would be
a. warm and rainy

QveST [7]2 years ago

4 0

Answer:

warm and rainy

Explanation:

because we expect a warm and rainy weather when the warm front is approaching

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Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

3 years ago

Read 2 more answers

How could an increase in industrial activity in developing nations contribute to global climate change?

netineya [11]

If there is an increase in industrial activity, that means that more heat will be dissipated to the atmosphere in the form of carbon dioxide. Industrialization requires fuel to keep the processes on the go. At the end of the pipeline, the combustion of fuel would result to carbon dioxide released to the atmosphere. That's how it is contributing to the global climate change through the greenhouse effect.

6 0

3 years ago

811.68 J of energy are used on a 95g chunk of unknown metal. Calculate the resulting temperature change if the specific heat of

Ivan

Answer:

16 °C

Explanation:

Step 1: Given data

Provided heat (Q): 811.68 J

Mass of the metal (m): 95 g

Specific heat capacity of the metal (c): 0.534 J/g.°C

Step 2: Calculate the temperature change (ΔT) experienced by the metal

We will use the following expression.

Q = c × m × ΔT

ΔT = Q/c × m

ΔT = 811.68 J/(0.534 J/g.°C) × 95 g = 16 °C

6 0

3 years ago

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

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5 0

1 year ago

Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)

Vladimir79 [104]

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

7 0

3 years ago