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2 years ago

Calculate how many moles are in 63.1 grams FeCl3.

Chemistry

1 answer:

Lynna [10]2 years ago

8 0

Answer: 1 grams FeCl3 = 0.0061650760770388 mole using the molecular weight calculator and the molar mass of FeCl3.

Explanation:

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Which atom has the least attraction for the electrons in a bond between that atom and an atom of hydrogen?

Helga [31]

Answer:

carbon

Explanation:

tbh im not sure just guessing

8 0

3 years ago

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

Sergeeva-Olga [200]

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

7 0

3 years ago

Help me out please!?

Sholpan [36]

Answer:

area a

Explanation:

6 0

3 years ago

A graduated cylinder is filled to the 40.00-mL mark with a mineral oil. The masses of the cylinder before and after the addition

castortr0y [4]

Answer:

Firstly, Let's experiment this !

Experiment 1 :

159.446g - 124.966g = 34.48g

34.48g = The mass of Mineral oil.

The density of the mineral oil = M/V = 34.48g/40mL = 0.862g/cm³.

Experiment 2 :

124.966 + 18.173 = 143.139 = The mass of solid + cylinder.

124.966 + 50.952 = 175.918 = The mass of solid + cylinder + Mineral water.

175.918 - 143.139 = 32.779 = The mass of added mineral oil.

Explanation:

Now we have to find the volume of the added mineral oil using the density from experiment 1.

V = 32.779g/0.862g/cm³ = 38.02668213mL

Since we found the volume of the solid, we then have to subtract the added mineral oil volume from the total volume from experiment 1.

Volume of solid = 40-38.02668213 = 1.97331787mL

Density of solid = 18.713g/1.97331787mL = 9.483013499g/cm^3

1.97331787 = (4/3)(3.14)r³

1.97331787*(3/4)(3.14) = .4713338861

.4713338861 = r ³

r = 0.7782328425158433

r = 0.78

Now that's our final answer ! r = 0.78

5 0

3 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of stomach acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

Hence, the volume of HCl neutralized is 1.25 mL

3 0

3 years ago