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3 years ago

2h2O---->2h2+O2

1 answer:

4 0

To solve this problem all you would have to do is to use the stoichiometric ratio of 1 mol of Oxygen would react with 2 mol of H2O, to find moles of water produced.

1 mol of O2 = 2 mol of H2O
20 mol of O2 = 40 mol of H2O.

The moles of water produced would be 40 moles.

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Empirical formula for 70.9% K and 29.1% S

aev [14]

I dont know sorry but good luck i know u can past

4 0

4 years ago

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole

Number of moles of $C_3H_8$ : $\frac{15 g}{144.1 g/mol}$ = 0.10 mole

Mass of CO : $\frac{15 g}{28.01 g/mol}$ = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of $CH_4$ : $\frac{4.3}{4.935}$ = 0.8713

Mole fraction of $C_3H_8$ : $\frac{0.10}{4.935}$ = 0.0202

Mole fraction of CO : $\frac{0.53}{4.935}$ = 0.107

Hence, composition of the stack gas are:

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

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8 0

3 years ago

3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g

hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

$\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol$

5 0

3 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.

The <u>first step</u> of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

4 years ago

Firlakuza [10]

Mixtures are made up of 2 or more pure substances. pure substance is just its self and a mixture is multiple substances.

6 0

2 years ago

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