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Anton [14]
3 years ago
8

Consider the chemical equations shown here.

Chemistry
1 answer:
brilliants [131]3 years ago
8 0

Answer:

ΔH for the the reaction NO(g) + O(g) ⇒ NO₂(g)  is ΔH= -304.1 kJ

Explanation:

<u>The complete question is:</u>

Consider the chemical equations shown here.

NO(g) + O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH= -198.9 kJ )

1.5 O₂(g) ⇒ O₃(g)  (ΔH= 142.3 kJ )

O(g) ⇒ 0.5 O₂(g) (ΔH= -247.5 kJ)

What is ΔH for the reaction shown below?

NO(g) + O(g) ⇒ NO₂(g)

Solution:

We have to use the Hess's Law: if a series of reagents react to give a series of products, the heat of reaction released or absorbed is independent of whether the reaction is carried out in one, two or more stages. That means enthalpy changes are additive.

NO(g)+ O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH₁= -198.9 kJ )

+

1.5 O₂(g) ⇒ O₃(g)  (ΔH₂= 142.3 kJ )

+

O(g) ⇒ 0.5 O₂(g) (ΔH₃= -247.5 kJ)

=

NO(g) + O₃(g) + 1.5 O₂(g) + O(g) ⇒ NO₂(g) + O₂(g) + O₃(g) + 0.5 O₂(g)

We remove the compounds that are in both members of the reaction:

NO(g) + O(g) ⇒ NO₂(g)

We only have to add the reactions so we add the value of each enthalpy change.

ΔH for the the reaction is given by:

ΔH= ΔH₁ + ΔH₂ + ΔH₃= -198.9 kJ +142.3 kJ -247.5 kJ= -304.1 kJ

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When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is r
butalik [34]

Answer:

There is 78.25g NaHCO3 required

Explanation:

Step 1: Balance the equation

2 NaHCO3 → Na2CO3 + CO2 + H20

For 2 moles of NaHCO3 consumed, there is produced 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of H2O.

Step 2: calculating moles of CO2

mass of Co2 = 20.5g

Molar mass of CO2 = 44.01 g/mole

moles of CO2 = 20.5 / 44.01 = 0.4658 moles

Step 3: Calculating moles of NaHCO3

Since we have for 1 mole CO2 produced, there is 2 moles of NaHCO3 consumed.

To calculate number of moles of NaHCO3, we have to multiply the number of moles of CO2, by 2.

⇒ 0.4658 x2 = 0.9316 moles

Step 4: Calculating the mass of NaHCO3

mass of NaHCO3 = moles of NaHCO3 x Molar mass of NaHCO3

mass = 0.9316 moles x 84g/ mole = 78.25g NaHCO3

There is 78.25g NaHCO3 required

4 0
3 years ago
Refer to attachment please this is one of my study questions and im stuck
Reil [10]
Percentage yield= actual yield/theoretical yield x100
So you would have to do-
15/22 x 100. Hope this helps!!
3 0
2 years ago
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina
liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0
3 years ago
An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques
fomenos
<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

A)

We know, each mole contains N_A= 6.023 \times 10^{23} atoms.

It is given that mass of one oxygen atom is m= 2.66\times 10^{-23}\ g.

Therefore, mass of one mole of oxygen, M=m\times N_A.

Putting value of n and N_A,

M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen=\dfrac{50}{16}=3.1 \ moles.

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3 0
3 years ago
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Answer:

Explanation:

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5 0
3 years ago
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