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3 years ago

Force causes

Chemistry

1 answer:

Rufina [12.5K]3 years ago

7 0

Answer:

is d or c

Explanation:

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Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrog

vitfil [10]

Answer:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Explanation:

To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.

Nitrite ion (NO₂⁻)

1 × N + 2 × O = -1

1 × N + 2 × (-2) = -1

N = +3

Nitrous oxide (NO)

1 × N + 1 × O = 0

1 × N + 1 × (-2) = 0

N = +2

Nitrate ion (NO₃⁻)

1 × N + 3 × O = -1

1 × N + 3 × (-2) = -1

N = +5

Ammonia (NH₃)

1 × N + 3 × H = 0

1 × N + 3 × (+1) = 0

N = -3

Nitrogen gas (N₂)

2 × N = 0

N = 0

The order of increasing nitrogen oxidation state is:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

7 0

4 years ago

If you rip a dollar bill in half is it qualified as 50 cents

omeli [17]

Yes, 50 pennies plus 50 pennies equals 100 pennies minus 50pennies equals 50 pennies.

7 0

3 years ago

PLEASE HELP WITH WORK INCLUDED

Mekhanik [1.2K]

I need more information?

7 0

3 years ago

Does indium react with acid?

Zinaida [17]

Yes it will react with acid.

4 0

3 years ago

Given that the concentration of myosin is 25pmol·L−1 and Rmax(Vmax)=208pmol·L−1·s−1 , determine the turnover number of the enzym

DedPeter [7]

Answer:

8.32 s⁻¹

Explanation:

Given that:

The concentration of myosin = 25 pmol/L

R_max = 208 pmol/L/s

The objective is to determine the turnover number of the enzyme molecule myosin, which has a single active site.

In a single active site of enzyme is known to be a region where there is binding of between substrate molecules, thereafter undergoing chemical reaction.

The turnover number of the enzyme is said to be the number of these substrate molecule which binds together are being converted into products.

The turnover number of the enzyme molecule of myosin can be calculated by the expression: $\dfrac{R_{max}}{concentration}$

⇒ $\dfrac{208 pmol/L/s}{25 pmol/L}$

= 8.32 s⁻¹

4 0

4 years ago