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Nonamiya [84]
2 years ago
9

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

l water and solution 2 has a volume of 175 ml and contains 7.75 g of zinc bromide. You mix the two solutions together in a large beaker. What is the bromide concentration in moles/L in the mixture
Chemistry
1 answer:
erma4kov [3.2K]2 years ago
8 0

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

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The combustion of a 0.4255 g sample of a compound containing carbon, hydrogen, oxygen and bromine produces 0.4961 g CO2 and 0.17
IgorC [24]

Answer: The empirical formula is C11HO3Br

Explanation: 1ST SAMPLE;

First, we need to get the number of moles;

Molar mass of C=12, O=16, H=1, Ag =107.9, Br=79.9

0.4961g CO2 x (1 mol CO2) / (44.0g CO2) = 0.0113 mol of Co2

Since one mole of CO2 is made up of 1 mole of C and 2 moles of O, and we have 0.0113 moles of CO2 in our sample, then we know we have 0.0113 moles of C in the sample. Now, we need to know the mass of C we have in the sample:

(0.0113 mol C) (12.011g C/1 mol C) = 0.136g C

Now we follow the same pattern above and do it for the hydrogen:

0.1777g H2O x (1 mol H2O) / (18g H2O) = 0.01 mol

Now, for the mass of H:

(0.01 mol H) (1g H/1 mol H) = 0.01g H

But this is for 1 mole of H. Whereas, H has 2 moles from the question, therefore it's equal to 0.01 x 2 = 0.02g H

Since we combusted 0.4255g of the sample, the missing mass will be from the bromine and oxygen since we have gotten masses of Carbon and Hydrogen.

Therefore missing mass = 0.4255 - (0.136+0.02) = 0.2695 of bromine and oxygen

2ND SAMPLE:

First, we need to get the number of moles;

0.1894 g AgBr x (1 mol AgBr) / (107.9 + 79.9g AgBr) = 0.001 mol of AgBr

Since AgBr is made up of 1 mole of Ag and 1 mole of Br each,

We can say that there's 0.001 mole of Br in this second sample.

Now looking at the first and second samples, lets set up a proportion to know the number of moles of Br in the first sample.

If 0.1523g of the second sample produced 0.001 mol of Br, therefore 0.4255g in the first sample will produce: (0.4255g x 0.001)/0.1523 = 0.0028mol of Br

Therefore the mass of Br in the first sample is: (0.0028 mol C) (79.9g Br/1 mol C) = 0.22372g Br

From the first sample, we saw that the sum of Br and Oxygen equals 0.2695

Therefore,the mass of oxygen is: 0.2695 - 0.22372 = 0.04578g of oxygen

Therefore to find number of moles of oxygen;

(0.04578g O x (1 mol O) / (16.0g O) = 0.0029 mol of oxygen

Overall, we have; C=0.0113 moles ;H=0.001 moles; Br= 0.0028moles and O= 0.0029

The smallest is 0.001. So to simplify this for the empirical formula, we divide each by 0.001 to get approximately C= 11, H=1, Br=3, O=3

Therefore the empirical formula is C11HO3Br

5 0
3 years ago
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KATRIN_1 [288]

Answer:

Yes Both A) KCL and D) HCL are correct...

3 0
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An atom of 125Sn has a mass of 124.907785 amu. Calculate the binding energy in MeV per atom. Use the masses: mass of 1H atom = 1
Ksenya-84 [330]

Answer:

1.06×10³ MeV/atom

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Also, atomic number of Sn = 5 0

Thus, the number of protons = 50

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

125 =  50 + Number of neutrons

Number of neutrons = 75

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,  

Calculated mass = (50*1.007825 + 75*1.008665) amu = 126.041125 amu

Given mass = 124.907785 amu

<u>Mass defect = Δm = |126.041125 - 124.907785| amu = 1.13334 amu </u>

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

<u>So, Energy = 1.13334*931.5 MeV/atom = 1.06×10³ MeV/atom</u>

3 0
3 years ago
How many grams of butane were in 1. 000 atm of gas at room temperature?
dimaraw [331]

The mass in grams of butane at standard room temperature is 53.21 grams.

<h3>How can we determine the mass of an organic substance at room temperature?</h3>

The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:

PV = nRT

  • Pressure = 1.00 atm
  • Volume = 22.4 L
  • Rate = 0.0821 atm*L/mol*K
  • Temperature = 25° C = 298 k

1 × 22.4 L = n × (0.0821 atm*L/mol*K×  298 K)

n = 22.4/24.4658 moles

n = 0.91556 moles

Recall that:

  • number of moles = mass(in grams)/molar mass

mass of butane = 0.91556 moles × 58.12 g/mole

mass of butane = 53.21 grams

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Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.

Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.

Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

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1 year ago
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