How many milliseconds (ms) are there in 3.5 seconds (2)

Oksana_A [137]

Answer:

the correct choice would be: 4. Reaction rates are not related to spontaneity.

Explanation:

The speed of reaction is independent of spontaneity, since it depends on other factors such as the amount of reagent.

The amount of reagent is one of the variables that most influences the speed of reaction, the greater the amount of reagent, the faster the reaction will be and thus the speed will be increased.

3 0

4 years ago

The value of ka for weak acid is to 4.0x10^-5 what is degree of dissociation of the acid in aqueous solution of concentration 0.

lakkis [162]

p

K

a

=

5.4

pH

=

3.7

Explanation:

p

K

a

is simply

−

log

10

(

K

a

)

. It is a constant at any temperature and does not depend on the molarity of the solution.

p

K

a

=

−

log

10

(

K

a

)

=

−

log

10

(

4.0

×

10

−

6

)

=

5.4Explanation:

3 0

3 years ago

Read 2 more answers

What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Etha

Arlecino [84]

Answer:

The % yield is 56.6 %

Explanation:

This is the reaction:

C₂H₄ + Cl₂ → C₂H₄Cl₂

Molar mass of ethylene gas: 28 g/m

Mol = mass / molar mass

140 g / 28 g/m = 5 moles

Ratio is 1:1, so 5 moles of ethylene produce 5 moles of dichloro ethane.

Molar mass of C₂H₄Cl₂ = 98.9 g/m

Mass of C₂H₄Cl₂ produced = 98.9 g/m . 5 m → 494.5 g

% yield reaction

(280 g / 494.5 g ) . 100 = 56.6%

6 0

4 years ago

Read 2 more answers

Select the correct electron configurations from the list below. You can refer to the periodic table for atomic numbers. Check al

vesna_86 [32]

What’s the list?.......

4 0

3 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago