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3 years ago

How do you balance the chemical equation NaOH + H2SO4 > Na2SO4 +H2O

1 answer:

5 0

Make sure there are the same number of atoms of each element on either side.
1) Check each one.
2) If one's out of balance, alter the equation to balance it and go back to 1)
3) When everything's balanced, you're finished!

We have 1 Na on the left and 2 Na on the right here. We'll need another NaOH to balance it:
2 NaOH + H₂SO₄ > Na₂SO₄ + H₂O
Now O is out of balance. There's 6 on the left and 5 on the right. We'll need more H₂O:
2 NaOH + H₂SO₄ > Na₂SO₄ + 2 H₂O
Fortunately H is in balance. S is also in balance here, so looks like we did it!

Need any more help?

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Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

6 0

4 years ago

For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

BlackZzzverrR [31]

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ( $\Delta G^{0}$ ) is given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

So, for the reaction of 1.57 moles of $NH_{3}$ , $\Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ$

As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.

6 0

3 years ago

Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?

lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

#SPJ4

7 0

2 years ago

When do plants use the process of cellular respiration

d1i1m1o1n [39]

Answer:

So the answer is every day

Explanation:

Plants respire at all times of the day and night because their cells need a constant energy source to stay alive.

8 0

4 years ago

Which elements did the Curies discover?

Scilla [17]

Answer:

B on edge

Explanation:

6 0

3 years ago