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jok3333 [9.3K]
3 years ago
12

In the reaction 4Al+_O2→2Al2O3 , what coefficient should be placed in front of the o2 to balance the reaction?​

Chemistry
2 answers:
deff fn [24]3 years ago
7 0

Answer:

yeep the answer is 3... i just took the k12 test rn

DIA [1.3K]3 years ago
4 0

Answer:

3

Explanation:

took the test

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Which volume measurement is expressed in four significant figures?
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Explanation:

                       

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Suppose that H2SO4 was used in the reaction instead of HCl. How many moles of NaOH would neutralize 1 mole of H2SO4?
arlik [135]

Answer: Hence, 2 moles of NaOH would neutralize 1 mole of H_2SO_4

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

One mole of hydrochloric acid neutralizes one mole of sodium hydroxide to give one mole of sodium chloride and one mole of water

When sulfuric acid is used in the place of HCl , 1 mole of sulfuric acid will neutralizes the 2 mole of sodium hydroxide and gives one mole of sodium sulfate and 2 moles of water. As we can see from the reaction:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

Hence, 2 moles of NaOH would neutralize 1 mole of H_2SO_4

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3 years ago
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What is the color of elemental silver? Explain.
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Use of fractional distillation ?
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3 0
4 years ago
A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
Charra [1.4K]

Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

pH = 0.119

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

5 0
3 years ago
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