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2 years ago

6

The following molecular scenes depict an important reaction in nitrogen chemistry (nitrogen is blue; oxygen is red): Write a bal

anced equation for this reaction. ____N2O5 ___NO2 + ___O2

1 answer:

alexandr1967 [171]2 years ago

3 0

Answer:

2 N2O5 4 NO2 + 1 O2

Explanation:

For the last one you dont really have to add the 1 but you can if its nedded,

since 3 divided by 3 is 1.

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Highly liquid lava forms wide, shield-like mountains. True False

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During the phase change from solid ice to liquid water, which bonds/forces are being weakened?

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A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

8 0

2 years ago

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV

allsm [11]

Answer:

The rate of change of the temperature is 0.0365 Kelvin per minute.

Explanation:

<u>Step 1</u>: Given data

ideal gas law: P*V = n*R*T

with P= pressure of the gas ( in atm) = 9.0 atm

with V= volume of the gass (in L) =12L

with n = number of moles = 10 moles

R = gas constant = 0.0821 L*atm* K^−1*mo^−1

T = temperature = TO BE DETERMINED

The volume decreases with a rate of 0.17L/min = dV/dT = -0.17

The pressure increases at a rate of 0.13atm/min = dP/dT

<u>Step 2:</u> The ideal gas law

P * [dV/dT] + V * [dP/dT] = nR * dT/dt

9 atm * (-0.17L/min) + 12L * 0.13atm/min = 10 moles * 0.0821 L*atm* K^−1*mo^−1 *dT/dt

0.03 = 0.821 * dT/dt

dT/dt = 0.03/0.821

dT/dt = 0.0365

Since the gas constant is expressed in Kelvin and not in °C, this means that <u>the rate of chagnge of the temperature is 0.0365 Kelvin per 1 minute.</u>

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Answer: 48.123 Mm

Explanation:

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2 years ago