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4 years ago

In the balanced reaction and using the mole ratio:

Chemistry

1 answer:

NikAS [45]4 years ago

4 0

12 moles are needed because stoichiometry is 1:2

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Why is NaHCO3 a base?

worty [1.4K]

NaHCO3 is a product of a strong base and a weak acid reaction. Thus it has weak basic properties.

HCO3- ion is actually amphoteric, which means it can act as a base or an acid. But it is weaker than a strong acid or a strong base.

<span>HCO3- is amphoteric meaning it acts both as a B.L. Acid and a B.L. Base.. which is why it's used to neutralize both acid and base spills in the lab.</span>

7 0

3 years ago

If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co

vladimir2022 [97]

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)

$\Delta n_g$ = (12 - 15)/2 = -3/2

We have;

$\Delta H = \Delta U + \Delta n_g\cdot R\cdot T$

Where R and T are constant, and ΔU is given we can write the relationship as follows;

$H = U + \Delta n_g\cdot R\cdot T$

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

$\Delta n_g$ = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

4 0

3 years ago

Draw the organic product of the following nucleophilic substitution reaction. Include all hydrogens atoms.

Sunny_sXe [5.5K]

<span>Answer: CH3-CH2-0-CH2-CH3 that is diethyl ether mechanism is SN2</span>

8 0

3 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

4 0

3 years ago

HELPP

Natasha_Volkova [10]

I’m pretty sure it’s D.increases the activation energy for a reaction.

4 0

3 years ago