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lesantik [10]
3 years ago
13

What are 3 atoms that have 4 valence electrons? ​

Chemistry
2 answers:
nadya68 [22]3 years ago
7 0
To know this, we look at the family numbers
I’m family 14, we know that each atom has 4 electrons
In family 14 there are the atoms:
Carbon
Silicon
Germanium
Tin
Lead
kow [346]3 years ago
7 0
It’s probably Carbon or Silcon
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What is the mass of 0.28 mole of iron?
vredina [299]

Answer:

Your answer should be 15.68 grams.

Explanation:

Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.

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A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor
Valentin [98]
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
 So X (the no.of moles of NaCl) = 0.027 m
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3 years ago
This picture shows a desert environment, which is home to many lizards, small mammals, and cacti
kotykmax [81]

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

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3 years ago
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fiasKO [112]

Answer:b

Explanation:sorry if I’m wrong

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