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4 years ago

What mass of hbr (in g) would you need to dissolve a 3.4 −g pure iron bar on a padlock?

Chemistry

1 answer:

vivado [14]4 years ago

5 0

Given:

Mass of pure iron (Fe) = 3.4 g

To determine:

Mass of HBr needed to dissolve the above iron

Explanation:

Reaction between HBr and Fe is

Fe + 2HBr → FeBr₂ + H₂

Based on the reaction stoichiometry-

1 mole of Fe reacts with 2 moles of HBr

# moles of Fe = mass of Fe/atomic mass of Fe = 3.4/56 g.mol⁻¹ = 0.0607 moles

Therefore # moles of HBr = 2*0.0607 = 0.1214 moles

Molar mass of HBr = 81 g/mole

Mass of HBr = 0.1214 moles * 81 g/mole = 9.83 g

Ans: Mass of HBR required is 9.83 g

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Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of s

maxonik [38]

Answer: The number of moles of $Na_2SO_4$ is 0.05 moles.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

$0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol$

The chemical reaction for the ionization of sodium sulfate follows the equation:

$Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)$

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = $\frac{1}{2}\times 0.1=0.05moles$ of sodium sulfate.

Hence, the number of moles of $Na_2SO_4$ is 0.05 moles.

8 0

3 years ago

PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP

anastassius [24]

Answer:

I believe the first one is correct but I think the second one would be the blood like substance. not a 100% sure sorry

6 0

3 years ago

Which statement best describes what occurs during a chemical reaction?

In-s [12.5K]

Answer:

bonds within the nucleus of reactants atoms are broken and rearranged to form new product atoms

3 0

3 years ago

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or

Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

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

Where, k is rate constant

So,

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

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

In percentage, the sample of C-4 remains = 0.7015 %

3 0

3 years ago

Complete the table by filling in the missing

navik [9.2K]

Answer:

A=a/ B=b/ C=b/ D=a/ E=B/ F=B/ G=c

Explanation:

Took one for the team, all answers above are right on edge 2020

8 0

3 years ago