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SVEN [57.7K]
2 years ago
7

4. The radius of a hydrogen atom is 37 pm (1pm 10"m). How many hydrogen atoms lined up side to side would it take to make 1.00 i

nch? (Hint start with 1.00 inch)
Chemistry
1 answer:
vfiekz [6]2 years ago
3 0

Answer:

V = 4/3 * 3.1416 * (37x10-10)3

V = 2.12x10-25 cm3

d = m/V

d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3

The difference in temperature, let's convert F to ºC:

ºC = -80-32/1.8 = -62.22 ºC

dT = -92.6 + 62.2 = -30.4 ºC

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Convert 5.4g of NaCl to moles
sammy [17]

Answer:

.0924 moles of NaCl

Explanation:

So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl

  1. You'll need to find the atomic mass of the compound NaCl to help you solve for moles
  • Sodium (Na) on the periodic table has a mass of 22.99
  • Chlorine (Cl) on the periodic table has a mass of 35.45

Add these two together----> 22.99 + 35.45 = 58.44

Now you can calculate for moles

<u>Written-out method:</u>

<u>5.4 grams of NaCl  |   1 mole of  NaCl              </u>

                                | 58.44 grams NaCl                = .0924 moles of NaCl

<u>Plug into calculator method:</u>

(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)

7 0
2 years ago
Looking at the name of an acid, how do you know if it only contains two elements or if it contains more than two?
Lisa [10]
Because it has two syllables
4 0
3 years ago
Read 2 more answers
The molecule shown is a
Pani-rosa [81]

c is the answer a benzene

8 0
3 years ago
Help name this, it’s chemistry
vova2212 [387]

Answer:

I believe it's

4-floro-5,6,8-Tribromo-3,3,7-trimethylnonane

5 0
2 years ago
You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2
avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).  

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0
3 years ago
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