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2 years ago

A sucrose packet contains 4.0g sucrose C12H22O11, how many moles of sucrose does the packet contain?

Chemistry

1 answer:

SpyIntel [72]2 years ago

8 0

Answer:

And we have to calculate the number of moles of sucrose present in a lb mass of sucrose: Moles of sucrose=454⋅g342.30⋅g⋅mol−1=1.33⋅mol .

Explanation:

Brainliest Answer Pls

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What common carboxylic acid does vinegar contain?

Dafna1 [17]

Acetic acid.
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7 0

3 years ago

PLEASE HELP! I really don’t understand! Thanks!

PIT_PIT [208]

http://century.rochester.k12.mn.us/cms/One.aspx?portalId=3086882&pageId=6133921

Hope this helps!

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6 0

3 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mars2501 [29]

Answer:

Qm = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0

3 years ago

Read 2 more answers

A sample of gas in which [h2s] = 3.50 m is heated to 1400 k in a sealed vessel. after chemical equilibrium has been achieved, wh

Bond [772]

The value of Kc for the thermal decomposition of H₂S is 2.2 x 10⁻⁴ at 1400 K:
2 H₂S(g) ↔ 2 H₂(g) + S₂(g)
initial 3.5 M 0 0
at equilibrium 3.5 M - 2x 2x x
Kc = [S₂][H₂]² / [H₂S]²
2.2 X 10⁻⁴ = x(2x)² / (3.5 - 2x)²
2.2 x 10⁻⁴ = 4 x³ / (3.5)² Assuming x <<<<< 3.5
x = 0.088
Thus [H₂S] = 3.324 M

6 0

3 years ago

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

6 0

3 years ago