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alex41 [277]
3 years ago
8

Potassium permanganate, KMnO, and glycerin, C3H5(OH)3, react explosively according to the

Chemistry
1 answer:
Finger [1]3 years ago
4 0

The volume of CO2 at STP =124.298 L

<h3>Further explanation</h3>

Given

Reaction

4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O

701,52 g of KMnO4

Required

volume of CO2 at STP

Solution

mol KMnO4 (MW=158,034 g/mol) :

mol = mass : MW

mol = 701.52 : 158.034

mol = 4.439

mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549

At STP 1 mol = 22.4 L, so for 5.549 moles :

=5.549 x 22.4

=124.298 L

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How many oxygen (O) atoms are in a molecule of C3H403?<br> O A. 1<br> O B. 10<br> O c. 4<br> O D. 3
sleet_krkn [62]

Answer:

your answer is 3

Explanation:

6 0
2 years ago
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If 0.500 mol of each of the following solutes is dissolved in 2.0 L of water, which will cause the greatest increase in the boil
serious [3.7K]

Answer is: D. Na2SO4.

b(solution) = 0.500 mol ÷ 2.0 L.

b(solution) = 0.250 mol/L.

b(solution) = 0.250 m; molality of the solutions.

ΔT = Kf · b(solution) · i.

Kf - the freezing point depression constant.

i - Van 't Hoff factor.

Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).

Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.

Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.

Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).

Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.

8 0
3 years ago
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4. When a turbine spins, it generates electrical energy and that is where it gets its name from.
Elena L [17]
If it’s hydraulic turbine then it’s potential and kinetic energy and if it’s a thermal process then heat energy from the fuel burnt runs the turbine
8 0
2 years ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-
hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

Second empirical formula is between Zn^{2+}(Positive ion) and O^{2-}(Negative ion)

Second Formula would be Zn_{2}O_{2}

Note : When the subscript are same they get cancel out, so Zn_{2}O_{2} would be written as ZnO

Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

Note- The subscript will be simplified and the formula will be written as NiO_{2}.

The empirical formula of four binary ionic compounds are : ZnF_{2}, ZnO, NiF_{4},NiO_{2}


8 0
3 years ago
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If the half-life of a radioactive substances is 590 million years and you have 40 atoms of it, how many half-lives will have pas
Nady [450]

Answer:

3

Explanation:

Applying,

2^{n'} = R/R'............... Equation 1

Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.

From the question,

Given: R = 40 atoms, R' = 5 atoms

Substitute these values into equation 1

2^{n'} = 40/5

2^{n'} = 8

2^{n'} = 2³

Equation the base,

n' = 3

3 0
3 years ago
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