All categories

3 years ago

Calculate how many valence electrons are in H2CO, N2, Cl2?

Chemistry

2 answers:

netineya [11]3 years ago

8 0

In H2CO, C is bonded to H, H and O. Write down the valence electrons of each element.2 H = 1x 2 = 2.C= 4O = 6Total: 12

Ket [755]3 years ago

6 0

<span>In H2CO, C is bonded to H, H and O. Write down the valence electrons of each element. 2 H = 1x 2 = 2. C= 4 O = 6 Total 12 </span>

You might be interested in

How many atoms of Kr (Krypton) are in a balloon that contains 2.00 mol of Kr? (4)

Svetllana [295]

Answer:

$atoms= 1.204x10^{24}atoms$

Explanation:

Hello!

In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:

$atoms=2.00mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms= 1.204x10^{24}atoms$

Best regards!

4 0

2 years ago

What would happen if the cell cycle stopped in a multicellular organism

oksano4ka [1.4K]

The organism would no longer grow.

5 0

2 years ago

A graduated cylinder was filled to 20.3 mL with water. A solid object weighing 73.05 g was immersed in the water, raising the me

victus00 [196]

The density of the solid object will be 2.63 g/mL

<h3>What is density?</h3>

Density of objects = mass/volume.

Recall that an object will always displace its own volume when placed in a liquid.

Volume of the solid object = Cylinder reading after immersing the object in the water - cylinder reading before immersing the object in the water.

= 48.1 - 20.4

= 27.8 mL

Mass of the solid object = 73.05 g

Density of the object = 73.05/27.8

= 2.63 g/mL

More on density can be found here: brainly.com/question/15164682

#SPJ1

3 0

8 months ago

Show the calculation of the mass (grams) of a 14.5 liter gas sample with a molecular weight of 82 when collected at 29°C and 740

mote1985 [20]

Answer:

In this conditions, the gaswll weight 46.74 g.

Explanation:

The idal gas law states that:

PV = nRT,

P: pressure = 740 mmHg = 0.97 atm

V: volume = 14.5 L

n: number of moles

R: gas constant =0.08205 L.atm/mol.K

T: temperature = 29°C = 302.15K

$n = \frac{PV}{RT} \\n = \frac{0.97x14.5}{0.082 x 302.15 } \\n = 0.57 mol$

1 mol gas ___ 82 g

0.57 mol gas __ x

x = 46.74 g

4 0

2 years ago

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL

dybincka [34]

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

Best regards!

5 0

2 years ago