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Alexandra [31]
3 years ago
9

C. Oxide is a simple radical why??​

Chemistry
1 answer:
Luda [366]3 years ago
7 0

Answer:

In chemistry, a radical is an atom, molecule, or ion that has an unpaired valence electron. ... In living organisms, the radicals super oxide and nitric oxide and their reaction products regulate many processes

Explanation:

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Convert 296 degrees Celsius to Kelvin.<br> A. 23 K<br> B. 552 K<br> C. 569 K<br> D. 550 K
katovenus [111]

Answer:

option c) 552 K is correct

Explanation:

we know that

0° C +273= 273 K

so

296° C+273=569 K

7 0
2 years ago
Read 2 more answers
If an object is red, what type of light is being absorbed by the material?
12345 [234]
All colors but red is being absorbed by the material.
5 0
3 years ago
When beryllium ion combines with carbonate ion, the numbers of each ion are??
Katyanochek1 [597]

Answer:

One of each

Explanation:

Be is in Group 2, so it loses its two valence electrons in a reaction to form Be²⁺ ions.

Carbonate ion has the formula CO₃²⁻.

We can use the criss-cross method to work out the formula of beryllium carbonate.  

The steps are

Write the symbols of the anion and cation.

Criss-cross the numbers of the charges to become the subscripts of the other ion.

Write the formula with the new subscripts.

Divide the subscripts by their highest common factor.

Omit all subscripts that are 1.

When you use this method with Be²⁺ and CO₃²⁻, you might  be tempted to write the formula for the beryllium carbonate as Be₂(CO₃)₂

However, you can divide the subscripts by their largest common factor (2).

This gives you the formula Be₁(CO₃)₁.

We omit subscripts that are 1, so the correct formula is  

BeCO₃

There is one Be²⁺ ion and one CO₃²⁻ ion in a formula unit of beryllium carbonate.

4 0
4 years ago
What is the percent of Heterozygous rabbits
xxMikexx [17]

Answer:

100 percent  heterozygous and 0 percent of white rabbits

Explanation:

6 0
3 years ago
Read 2 more answers
10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and
Shalnov [3]

Answer:

The density is 3g/L

Explanation:

The reaction that occurs in the vessel is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

10,0g of NH₃ are:

10,0g * \frac{1mol}{17,031g} = 0,587 moles

6,50 g of O₂ are:

6,50g * \frac{1mol}{32g} = 0,203 moles

For a complete reaction of O₂ there are necessaries:

0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}

O₂ is limiting reactant. The excess moles of NH₃ are:

0,587 - 0,163 = <em>0,424 moles of NH₃</em>

These moles are:

0,424mol * \frac{17,031g}{1mol} = <em>7,22g of NH₃</em>

Knowing O₂ is limiting reactant, mass of NO and H₂O are:

0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} = <em>4,87g of NO</em>

0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} = <em>4,39g of H₂O</em>

The total mass is: 7,22g + 4,87g + 4,39g = 16,48g ≡ <em>16,5g </em>

<em>-</em><em>The same mass add in the first. By matter conservation law-</em>

As vessel volume is 5,50L, density is:

16,5g/5,50L = <em>3g/L</em>

I hope it helps!

7 0
4 years ago
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