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Oksanka [162]
3 years ago
9

Matter that is composed of two or more different elements chemically combined in a fixed proportion is classified as...

Chemistry
1 answer:
Lilit [14]3 years ago
4 0
It's called a compound because different elements are held together by a chemical bond.
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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
True or false: Fungi are able to carry out photosynthesis, similar to plants
QveST [7]

Answer:

false

Explanation:

7 0
2 years ago
Read 2 more answers
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
3 &amp; 4 Help pleaseeeee
kvasek [131]
3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...
4 0
3 years ago
The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate
Hitman42 [59]

<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

7 0
3 years ago
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