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1 year ago

Write the chemical formula for tin(ii)sulfite.. Please show working thank you

1 answer:

6 0

Accordingly, the majority of the bonds in the molecule, tin (ii) sulfide, are covalent, accounting for around 90% of the bonds. An extremely thin anode interlayer is used in organic photovoltaics by the compound tin(II) thiocyanate, Sn(SCN)2.

Stannic sulfide, tin disulfide, and tin bis(sulfanylidene) are examples. EC number: 215-252-9; CAS number: 1315-01-1. A chemical compound is tin(II) sulfate. SnSO4 is its atomic number. Tin and sulfate ions can be found in it. The chemical element tin has the atomic number 50 and the letter Sn for its symbol. At room temperature, tin is solid and is categorized as a post-transition metal. In order to create Tin dioxide and two molecules of hydrogen gas, Tin combines with two molecules of water, which are really in the form of steam. A reminder: Acids and alkalis corrode tin.

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The concentration of hydrogen peroxide solution can be determined by

max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- = 0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution = 5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0

3 years ago

The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium

Veronika [31]

Answer:

Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]

Explanation:

According to Brönsted-Lowry acid-base theory:

An acid is a substance that donates H⁺.
A base is a substance that accepts H⁺.

When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.

CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)

The basic equilibrium constant (Kb) is:

Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]

3 0

3 years ago

In the redox conversion of Ni2+ to NiO4−, the oxidation number of Ni goes from (−2, 0, +2) to (−1, +1, +7, +9). Recall that the

inessss [21]

Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7

7 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

Which of the data sets qualitative data?

Kaylis [27]

Answer:

Uh, the second one?

Explanation:

Try to restate the question please.

3 0

3 years ago

Read 2 more answers