All categories

3 years ago

How many grams of sodium acetate are in solution in the third beaker?

Chemistry

1 answer:

Kipish [7]3 years ago

6 0

Answer:

46g of sodium acetate.

Explanation:

The data is: Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL there are 46g of sodium acetate in solution. The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

You might be interested in

The modern atomic model includes electrons orbiting the nucleus. which of the following parts of dalton's atomic model was dispr

lana66690 [7]

Atom can never be divided into smaller particles is the answer was disproved

7 0

3 years ago

John rode 3150 m at an average speed of 350 m/min.If he had ridden at average of 375 m/min instead. How much sooner would it hav

Svet_ta [14]

Explanation:

Given that:

Distance traveled = 3150m

Average speed = 350m/min

Suggested speed = 375m/min

Unknown:

Time = ?

Solution:

Speed is the rate of change of distance with time.

it is mathematically expressed as;

speed = $\frac{distance }{time}$

Initial time using speed 350m/min can be calculated by making time the subject of the formula;

time taken = $\frac{distance }{speed}$

time taken at speed of speed of 350m/min = $\frac{3150}{350}$ = 9min

to seconds = 9 x 60 = 540 seconds

time taken at speed of 375m/min = $\frac{3150}{375}$ = 8.4min

to seconds = 8.4 x 60 = 504 seconds.

It could have taken 504 seconds sooner saving 36 seconds.

learn more:

Speed brainly.com/question/10177389

#learnwithBrainly

7 0

3 years ago

Write a balanced equation for the complete combustion of 2,3-dimethylbutane. use the molecular formula for the alkane (c before

Nuetrik [128]

2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O

Step 1. Write the condensed structural formula for 2,3-dimethylbutane.

(CH_3)_2CHCH(CH_3)_2

Step 2. Write the molecular formula.

C_6H_14

Step 3. Write the unbalanced chemical equation.

C_6H_14 + O_2 → CO_2 + H_2O

Step 4. Pick the most complicated-looking formula (C_6H_14) and balance its atoms (C and H).

1C_6H_14 + O_2 → 6CO_2 + 7H_2O

Step 5. Balance the remaining atoms (O).

1C_6H_14 + (19/2)O_2 → 6CO_2 + 7H_2O

Oops! Fractional coefficients!

Step 6. Multiply all coefficients by a number (2) to give integer coeficients..

2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O

4 0

3 years ago

Read 2 more answers

Help me please!

Vlad1618 [11]

Answer:

C. 108 grams/100g of H2O

Required number is the vertical coordinate of the intersection point of a line at 60°C with the graph of the KNO₃.

8 0

2 years ago

Read 2 more answers

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!

5 0

3 years ago