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Afina-wow [57]
3 years ago
14

What is the answer to this question?

Chemistry
1 answer:
scZoUnD [109]3 years ago
8 0

Answer:

DUDE ITS THE THIRD ANSWER FROM THE TOP

....... HOPE THIS HELPS ✌

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Solid sodium reacts with an iron (III) chloride solution to produce solid iron and sodium chloride solution. State the skeletal
AVprozaik [17]

Answer:

Na + FeCl3 -----NaCl + Fe

4 0
3 years ago
Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) +
siniylev [52]

Answer:

5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.

Explanation:

The balanced reaction:

CO (g) + 2 H₂ (g) -> CH₃OH (l)

By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:

  • CO: 1 mole
  • H₂: 2 moles
  • CH₃OH: 1 mole

Being the molar mass of each compound:

  • CO: 28 g/mole
  • H₂: 1 g/mole
  • CH₃OH: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CO: 1 mole* 28 g/mole= 28 grams
  • H₂: 2 moles* 1 g/mole= 2 grams
  • CH₃OH: 1 mole* 32 g/mole= 32 grams

Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:

  • If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

mass of carbon monoxide=\frac{6000 grams of methanol*28 grams of carbon monoxide}{32 grams of methanol}

mass of carbon monoxide= 5250 grams= 5.25 kg

If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

mass of hydrogen=\frac{6000 grams of methanol*2 grams of hydrogen}{32 grams of methanol}

mass of hydrogen= 375 grams

<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>

8 0
3 years ago
What is the final product of the following sequence of reactions? A. A B. B C. C D. D 21. Which of the following reacts the fast
Gala2k [10]

Answer:

Correct answer is Option A:

A. ethylene oxide (oxirane)

Explanation:

It reacts faster because it leads to a tertiary cation.

5 0
4 years ago
What mass of zinc oxide would be produced by the thermal decomposition of 375 grams of zinc carbonate?
Llana [10]
Zinc carbonate has the chemical formula : ZnCO3 and the equation for thermal decomposition is:
ZnCO3 .................> ZnO + CO2
1 mole of ZnCO3 produces one mole of ZnO

From the periodic table:
molar mass of zinc = 65.38 gm
molar mass of carbon = 12 gm
molar mass of oxygen = 16 gm

molar mass of ZnCO3 = 65.38 + 12 + 3(16) = 125.38 gm
molar mass of ZnO = 65.38 + 16 = 81.38 gm

125.38 gm of ZnCO3 produces 81.38 gm of ZnO, therefore:
mass of ZnO in 375 gm = (375 x 81.38) / 125.38 = 243.4 gm

Based on the above calculations, the correct answer is 243 grams


6 0
4 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

4 0
3 years ago
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