Explain why grinding the solid increases the rate of solution.

The acetoacetic ester synthesis is a method for preparing methyl ketones from alkyl halides.

ad-work [718]

The answer is A) true

5 0

2 years ago

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Bond [772]

Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2) $^{t/t(1/2)}$ where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days

5 0

3 years ago

11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J

Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

1

∘

C

.

In your case, aluminium is said to have a specific heat of

0.90

J

g

∘

C

.

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

1

∘

C

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

1

∘

C

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

1

∘

C

, you'd have to provide it with

1 gram



0.90 J

+

1 gram



0.90 J

+

...

+

1 gram



0.90 J



10 times

=

10

×

0.90 J

However, you don't want to increase the temperature of the sample by

1

∘

C

, you want to increase it by

Δ

T

=

55

∘

C

−

22

∘

C

=

33

∘

C

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

1

∘

C



10

×

0.90 J

+

1

∘

C



10

×

0.90 J

+

...

+

1

∘

C



10

×

0.90 J



33 times

=

33

×

10

×

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

33

∘

C

will be

q

=

10.0

g

⋅

0.90

J

g

∘

C

⋅

33

∘

C

q

=

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

q

=

m

⋅

c

⋅

Δ

T

, where

q

- the amount of heat added / removed

m

- the mass of the substance

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0

3 years ago

Bandingkan kesan penyusuan susu ibu dengan penggunaan susu formula terhadap<br>perkembangan bayi.

DedPeter [7]

Answer:

kau kje nk senang je kan . gi cari jawapan dlm buku la

Explanation:

aku x nk jwb . benda bole cri kt buku . ni la anak zmn skg ni . semua nk mudah

5 0

2 years ago

An acid sample of an unknown concentration is contained in a erlenmeyer flask. Which technique below would be best suited to ana

Ivenika [448]

Answer:

Titration

Explanation:

The best technique which can be used to determine the number of moles of the HCl in the sample is titration.

The given amount of HCl solution must be titrated with known concentration of the base like NaOH.

The volume of NaOH required must be noted also.

According to the reaction,

$NaOH+HCl\rightarrow NaCl+H_2O$

At equivalence point

Moles of $HCl$ = Moles of $NaOH$

Considering:-

Moles of $HCl=Molarity_{NaOH}\times Volume_{NaOH}$

Thus, in this way, moles of HCl can be determined.

4 0

3 years ago