PLSSS HELP. And there’s also at the bottom of the picture another option if u didn’t see it

A chemist adds of a sodium thiosulfate solution to a reaction flask. Calculate the mass in grams of sodium thiosulfate the chemi

schepotkina [342]

This is a incomplete question.The complete question is:

A chemist adds 180.0 ml of a 1.77 mol/L of sodium thiosulfate solution to a reaction flask. Calculate the mass in grams of sodium thiosulfate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: 50.4 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of sodium thiosulfate solution = 1.77 M

Volume of sodium thiosulfate solution = 180.0 mL = 0.1800 L

Putting values in equation 1, we get:

$\text{Moles of sodium thiosulfate}={1.77}\times{0.1800}=0.319moles$

Mass of sodium thiosulfate = $moles\times {\text {Molar mass}}=0.319moles\times 158.11g/mol=50.4g$

Thus 50.4 g of sodium thiosulfate the chemist has added to the flask.

6 0

3 years ago

Complete the passage about metallic bonding.

Burka [1]

Answer:

some bonds are broken and new ones are formed. Now you are ready to learn more about those bonds. Chemical bonds are attractions between atoms. They are simply attractive forces (between the + nucleus of one atom and the - electrons of a neighboring atom) that hold groups of atoms together and make them function as a unit.

7 0

3 years ago

Barium (Ba) is in Group 2. Recall that atoms in Group 1 lose one electron to form ions with a 1+ charge. What type of ion does b

Marizza181 [45]

Barium is in group 2A, it looses 2 electrons to form the calcium ion BA2, i’ll send a picture of some of my notes, hopefully it helps:))

6 0

3 years ago

A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of he antacid was added to 200 m

SSSSS [86.1K]

Answer:

a) 8.33 ml of the original stomach acid is neutralized

b) 191.67 ml of the stomach acid was neutralized

c) 249.68 ml acid would be neutralized by the original tablet

Explanation:

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid

8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

b) how much stomach acid was neutralized y the 4.3628 g tablet?

It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)

This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

4.3628 g antacid is equivalent to 191.67 ml acid ( this is the answer for question 2)

5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid

3 0

3 years ago

Read 2 more answers

Simple distillation diagram

tensa zangetsu [6.8K]

Answer:ccccccccccccccccccccccccccccccccccccccc

Explanation:

8 0

3 years ago