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4 years ago

Which compound contains only nonpolar covalent bonds? O 02 0 CO2 O H20 O NO3

Chemistry

2 answers:

Aneli [31]4 years ago

7 0

Answer:

A o2

Explanation:

edge 2020

Alenkinab [10]4 years ago

3 0

Answer:

Explanation:

O2 contains nonpolar covalent bonds.

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The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay

pickupchik [31]

Answer:

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

Explanation:

The unbalanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z = 27 + 1 = 28

Your nuclear equation becomes

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}$

Element 28 is nickel, so the balanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

7 0

3 years ago

(a) 7.56 0.375 +14.3103<br>

Musya8 [376]

Answer:Use the Law of Sines to solve the triangle ABC if it exists: a) A= 68.41°, B = 54.23”, a = 12.75 ft. b = b = b = a sub ź111347.

Explanation:

4 0

3 years ago

Will someone actually help me and not just something random

Novosadov [1.4K]

Answer:

Explanation:

8 0

4 years ago

An atom has radius of 227 pm and crystallizes in a body-centered cubic unit cell. What is the volume of the unit cell

sdas [7]

We will see that the volume of the unit cell is 144,070,699.06 pm^3

<h3>How to get the volume of a body-centered cubic unit cell?</h3>

In a body-centered cubic unit cell, the side length of the cube is given as:

$S = \frac{4}{\sqrt{3} } *R$

Where R is the radius of the atom.

And the volume of a cube is the side length cubed, then we can see that the volume of our cube will be:

$V = S^3 = (\frac{4}{\sqrt{3} }*227pm)^3$

Solving that we get:

$V = (\frac{4}{\sqrt{3} }*227pm)^3 = 144,070,699.06pm^3$

This is the approximated volume of the unit cell.

If you want to learn more about unit cell structures, you can read:

brainly.com/question/13110055

5 0

2 years ago

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Musya8 [376]

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$ :

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

= $\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$ :

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{11 mol}{45 mol}\times 100=24.44\%$

7 0

4 years ago