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Assoli18 [71]
2 years ago
6

Which of the following is a precipitation reaction? a) Zn (s) + 2 AgNO3 (aq) → 2 Ag (s) + Zn(NO 3) 2 (aq) b) 2 LiI (aq) + Hg 2(N

O3)2 (aq) → Hg 2I2 (s) + 2 LiNO3 (aq) c) HCl (aq) + KOH (aq) → KCl (aq) + H 2O (l) d) NaCl (aq) + LiI (aq) → NaI (aq) + LiCl (aq) e) None of the above.
Chemistry
1 answer:
lawyer [7]2 years ago
4 0

Answer: 2LiI(aq)+Hg_2(NO_3)_2(aq)\rightarrow Hg_2I_2(s)+2LiNO_3(aq)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

Precipitation is a type of displacement reaction in which one the products is formed in the solid state.

2LiI(aq)+Hg_2(NO_3)_2(aq)\rightarrow Hg_2I_2(s)+2LiNO_3(aq)

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What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?
zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

V2 = 500 mL

V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL

3 0
3 years ago
I don't really understand this worksheet question.​
Citrus2011 [14]
25/2 and 96/X
CROSS MULTIPLY.

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Measurements of the heights of various plants in an experiment are called
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Option 1 Data

5 0
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Someone anyone give me tips to finish online homework because i have 30 assignments to do
MrMuchimi

Answer:

Ok! I have four tips!

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6 0
3 years ago
Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years
Sliva [168]

Answer : The time passed in years is 2.83\times 10^3\text{ years}

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

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

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = 71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}

t=2831.00\text{ years}=2.83\times 10^3\text{ years}

Therefore, the time passed in years is 2.83\times 10^3\text{ years}

8 0
2 years ago
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