3 years ago

I COULD REALLY USE SOME HELP! I WILL RATE BRAINLIEST!

Chemistry

1 answer:

Travka [436]3 years ago

3 0

0.000169 mol/g citric acid

You might be interested in

Convert 6.33×10−6 cg to nanograms.

tensa zangetsu [6.8K]

5.73e+8 hope this help

5 0

3 years ago

Write formulas for the binary ionic compounds formed between the following elements: a. Potassium and iodine b. Magnesium and ch

lianna [129]

i think it's B Magnesium and chlorine

5 0

3 years ago

Explain why the level of greenhouse gasses in the atmosphere is rising.

Murljashka [212]

Is a gas in the greenhouse that absorbs and emits radiation within the thermal infrared range.

3 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

3 years ago

Puzzle: A man stands on a road in Iowa looking up at the night sky. On one night he sees the constellation Orion. But on another

nasty-shy [4]

Answer:

altho I don't believe in supper stision I would say that Scorpius has killed Orion.

5 0

2 years ago