All categories

2 years ago

In the world of chemistry, a 'mole' is a number of something... a very large number of something.

Chemistry

1 answer:

kipiarov [429]2 years ago

3 0

Given:

No of atoms present= 8.022 x 10^23 atoms

Now we know that 1 mole= 6.022 x 10^23 atoms

Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.

Number of moles

= 8.022 x 10^23/6.022x 10^23

=1.3 moles.

Hence we have 1.3 moles present.

You might be interested in

HELP ME ON NUMBER 8 ASAP

Varvara68 [4.7K]

Answer:

2. hope it helps. please follow me I'm new here

3 0

3 years ago

Numerator and denominator are Equal to one another? I don’t get the concept

Alekssandra [29.7K]

It's like saying that if 1 meter is equal to 100 cm,

$\frac{100cm}{1m}$

$\frac{1m}{100cm}$

You can base the answers from the chart

5 0

2 years ago

When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balan

Damm [24]

Answer:

a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) 67.6%

Explanation:

a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).

Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.

In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:

Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.

First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:

1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄

Transforming to mass (mass = moles * molar mass):

331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄

x ------------------------- 12.38

By a simple direct three rule:

162x = 4100.256

x = 25.3 g of Pb(NO₃)₂

This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:

1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)

Transforming to mass:

331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)

11.39 g ------------------- y

By a simple direct three rule:

331.2 y = 3681.248

y = 11.115 g

The yield is the mass formed divided by the stoichiometric result multiplied by 100%:

yield = (7.52/11.115)*100% = 67.6%

7 0

3 years ago

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to

Natalija [7]

Answer:

1- 3 Moles of CO2

2- 132 g of CO2

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g O2 ⇒ 132 g CO2

128 g O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

44 g C3H8 ⇒ 132 g CO2

132,33 g C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

44 g C3H8 ⇒ 72 g H2O

132,33 g C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

3 0

2 years ago

draw two lewis structures for a compound with the formula c2h6o and compare the boiling points of the two molecules

Volgvan

Answer:

See explanation and images attached

Explanation:

There are two compounds that could have molecular formula of C2H6O, they are ethanol and methoxy methane (trivial name: methyl ether). The structures of the two compounds are shown in the image attached as obtained from quora.

The boiling point of methoxymethane is -24.8°C while that of ethanol is about 78.5°C. The reason for the high boiling point of ethanol is the presence of intermolecular hydrogen bonding leading to a stronger intermolecular interaction compared to methoxymethane molecules held together only by very weak dispersion forces.

4 0

2 years ago