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iVinArrow [24]
3 years ago
14

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Chemistry
2 answers:
iogann1982 [59]3 years ago
4 0

Answer:

my big fhat black calwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwlk

viktelen [127]3 years ago
3 0

Answer:

HmmmmmmmmmmI think the answer is 69

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Fill in the blanks and balance with the correct coefficient​
MrRissso [65]

\huge\mathrm{  \underline{Answer}}࿐

The Balanced equation will be :

\mathrm{ \boxed2Fe +  \boxed3Cl_2  \rightarrow  \boxed2FeCl_3}

The Coefficients are :

  • Fe - 2

  • Cl_2 - 3

  • FeCl_3 - 2

_____________________________

\mathrm{ \#TeeNForeveR}

4 0
3 years ago
Solid sodium hydroxide <br> NaOH<br> decomposes into gaseous water and solid sodium oxide .
kari74 [83]
What about it do you want answered
8 0
3 years ago
I am a nonmetal.I am in the Oxygen family and in row 3.I have 6 valence electrons.I am yellow and have a stinky smell.Who am i
Dimas [21]

Answer:

sulfur

Explanation:

In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.

It is present in oxygen family.

It has six valance electrons.

Its atomic number is 16.

Its atomic weight is 32 amu.

The electronic configuration of sulfur is given below,

S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴

We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )

Sulfur is used in vulcanisation process.

It is used in bleach and also as a preservative for many food.

it is used to making gun powder.

8 0
3 years ago
Why is Anton Von Leeuwenhoek important in the cell theory?​
vesna_86 [32]

Answer:

He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.

Hope this helps :)

4 0
3 years ago
Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c
LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass  of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H =  4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:  

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%  

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0
3 years ago
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