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Helga [31]
3 years ago
14

Table salt is the compound sodium chloride (NaCl). Use the periodic table to find the molar mass of NaCl. g/mole

Chemistry
2 answers:
lord [1]3 years ago
7 0
From the periodic table, you will find that:
mass of Cl = <span>35.453 grams
mass of Na = </span><span>22.989769

Now, since NaCl is formed from the combination of one mole of sodium from one mole of chlorine, therefore, the molar mass of NaCl can be calculated as followed:
molar mass of NaCl = </span>35.453 + 22.989769 = 58.442769 gram/mole
lbvjy [14]3 years ago
6 0

58.44 to round it up. I don't believe E2020 takes the full answer. The other person is correct though.

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Unevenly

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Fresh water is distributed unevenly in both time and space.
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2 years ago
Of the elements: b, c, f, li, and na. the element with the smallest ionization energy is
love history [14]
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4 years ago
I need help please with this chemistry work ​
Marysya12 [62]

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8 0
3 years ago
Two Balloons one filled with hydrogen gas and second filled with neon gas. what gas should be used in an experiement to identify
dalvyx [7]

Answer:

Oxygen

Explanation:

If two balloons are filled with hydrogen gas and helium gas respectively, then we want to identify what gas is in each balloon, we have to do so by exposing the both balloons to flame in an oxygen atmosphere.

Hydrogen combines with oxygen in the presence of a flame with quite a loud sound and the flame is sustained but when a flame is brought near helium gas in a balloon, the gas will only make a little sound when exposed to the flame and extinguish the flame.

The reason for the explosion of the gas in the hydrogen balloon is that combustion of hydrogen gas is exothermic. The heating up of surrounding air molecules leads to a sudden explosion.

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8 0
3 years ago
Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. n_{CuO}= \frac{4}{80}= 0,05 (mol)

⇒n_{CuCl2}= n_{CuO}=0,05 mol

⇒m_{CuCl2}= 0,05×135=6,75 (g)

c. n_{HCl}=2× n_{CuO}=0,1 (mol)

⇒m_{HCl}= 0,1×36,5= 3,65 (g)

⇒m_{dd HCl}= \frac{m_{HCl}}{10}×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }×100=\frac{6,75}{36,5+4} ×100≈ 16,67%

8 0
3 years ago
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