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Andru [333]
3 years ago
10

What is the gram formula mass of Fe(NO3)3

Chemistry
1 answer:
olga55 [171]3 years ago
6 0
242 g/mol...... hope it helps :)
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If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca
Jlenok [28]

Answer:

\boxed{\text{8 da}}

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:

\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}

\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}

3 0
3 years ago
3 points
vfiekz [6]

Answer:

density=6.74g/ml

:320g÷47.5ml

d=6.74g/ml

thank you

<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>

8 0
3 years ago
25.0 cm3 of a solution of sodium hydroxide required 18.8 cm3 of 0.0500 mol/dm3 H2SO4
marshall27 [118]

Answer:

i think so

Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O

1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M

Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol

7 0
3 years ago
Current is applied to a molten mixture of AgF , FeCl2 , and AlBr3 . What is produced at each electrode? STRATEGY Rank the cation
ratelena [41]

Answer:

Cathode: Ag

Anode: Br₂

Explanation:

In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:

Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V

Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V

Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V

As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.

For the anode an oxidation must occurs, so the reactions for the nonmetals are:

F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V

Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V

Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V

For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.

5 0
3 years ago
help, i have a huge honors chem test that’s on like 3 or 4 chapters and i need to know how to do all the math but i don’t and i
meriva

Answer:

you should study.

Explanation:

it depends on when the exam is, but remember to make sure you study well for it and dont rely on studying it all the night of the exam.

5 0
3 years ago
Read 2 more answers
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