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2 years ago

10

Which of the following will result in a chemical change?

2 answers:

krok68 [10]2 years ago

8 0

It would be /A/....because a chemical change is somthing that cant NOT be brought back to its areganal proproties.coal will never be able to be bavk the way it was!!

Finger [1]2 years ago

5 0

A. Burning coal in a furnace. This is because when you burn coal, you cant get back to the reactants substance. Burning is a chemical change.

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A chemist performs the same tests on two homogeneous white crystalline solids, A and B. The results are shown in the table below

Nookie1986 [14]

Solid A contains only Covalent bonds and solid B only contains Ionic bonds. I know this because the appearance, solubility, and conductivity of solid A describes Covalent bonds.

Explanation:

Try this, but idk if it will be marked as correct.

7 0

3 years ago

Which half-reaction correctly represents reduction?

Neko [114]

Answer: B. Ca2+ + 2e- ---------> Ca

Explanation:

Reduction involves the gaining of electrons as well as a decrease in the oxidation state of the atom or ion.

In this case; Ca2+ gains two electrons and the oxidation state is reduced from +2 to 0.

Ca2+ + 2e- ---------> Ca

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2 years ago

Pb(NO3)2 + NH4Cl = PbCl2 + NH4NO3<br> how do i balance this??

Elden [556K]

Answer:

We have NH 4 and that's called the ammonium ion it also stays together.

Explanation:

7 0

2 years ago

How many moles of na are needed to produce 4 moles of nacl in this reaction? 2na + cl2 → 2nacl?

never [62]

14...........................

4 0

2 years ago

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago