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Consider this chemical reaction. What chemical is reduced?<br> CH3OH + NAD --> CH2O + NADH

nikklg [1K]

Answer:

CH3OH and NADH

Explanation:

The given chemical reaction is an redox reaction in which reduction and oxidation take place.

In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.

In the given redox reaction: CH3OH + NAD --> CH2O + NADH

NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.

So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.

5 0

3 years ago

How many elements compose the majority of rock-forming minerals?

PSYCHO15rus [73]

10 elements compose the bulk of rock forming minerals

4 0

3 years ago

What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!

Lemur [1.5K]

The answer to your question is nitrogen dioxide

8 0

3 years ago

The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$ .

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of $NiSO_4$ .

Molar mass of $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$ :

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$ :

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$ :

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$

3 0

3 years ago

Enter the chemical formula of a binary molecular compound of hydrogen and a Group 6A element that can reasonably be expected to

NARA [144]

Answer:

Any binary molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

Explanation:

Going down in a group increases the atomic radius and a greater atomic radius implyes greater ionic radius.

When ionization takes place in these compounds they yelds protons (hidrogen ion) and an lewis base (anion). The greater the ionic radius the greater its stability, thus the periodic tendency is increaing the acidity of binary hidrogen compounds when going down a group. On the other hand going up a group decreases acidity, so any molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

3 0

3 years ago