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xxTIMURxx [149]
3 years ago
11

H₃C-Ç=CCH₃сна сHз (what is the name of this chemical formula)​

Chemistry
1 answer:
algol [13]3 years ago
3 0

Answer:2,3 -dimethyl-2-butenr

Explanation:

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Which of the following is the Lewis structure for CO2?
velikii [3]

Answer:

The CO2 Lewis structure has two double bonds going from carbon to the oxygen atoms. According to the octet rule, each oxygen atom needs to bond twice and the carbon atom needs to bond four times.

4 0
2 years ago
4 HF(g)+SiO2(s) → SiF4(9)+2 H2O(9) <br> Is the Si oxidized or reduced?
Airida [17]

Answer:

Si is reduced since it loses the oxygen atom

8 0
3 years ago
There are 7 named classes of hazardous materials.<br> O True<br> O False
Anika [276]
False

there is actually 9
7 0
2 years ago
Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0
3 years ago
Calculate the molarity of a solution with 233.772g sodium chloride dissolved in 2,000mL of water
LekaFEV [45]

Answer:

The molarity is 2M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl

<em>A solution molar--> moles of solute in 1 L of solution:</em>

2 L-----4 mol NaCl

1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M

3 0
3 years ago
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