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2 years ago

8

We have a sample of water vapor which cools from 150 C to 125 C what happens to the motion of the molecules during this time of

cooling

1 answer:

Fofino [41]2 years ago

7 0

Any options?????????

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Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is

Korolek [52]

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

$\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2$

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

<u>Method 1: using the </u><u>mass of glucose</u>

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= $\frac{10}{9}$ mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

= $2(\frac{10}{9} )$

= $\frac{20}{9}$ mol

Mass of ethanol

= $\frac{20}{9} \times[2(12)+6+16]$

= $\frac{20}{9}(46)$

= 102 g (3 s.f.)

<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= $\frac{977}{440}$ mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= $\frac{977}{440}$ mol

Mass of ethanol formed

= $\frac{977}{440} \times[2(12)+6+16]$

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

brainly.com/question/1685725

8 0

1 year ago

Solve for y in the following problem: 5.3 x 10- (y)(2y)

weqwewe [10]

Answer:

The value of y = 5.1478

Explanation:

The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.

The given equation: 5.3 x 10- (y)(2y) = 0

⇒ 53 - 2y² = 0

⇒ 2y² = 53

⇒ y² = 53 ÷ 2 = 26.5

⇒ y = √26.5 = 5.1478

8 0

3 years ago

describe an example of how humans used selective breeding in plants,and explain how that has benefited us

Crank

The plants give out oxygen so for us humans will live

6 0

3 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

2 years ago

Write word equation for reaction between sodium and sulphur

White raven [17]

Answer:

2Na + S —> Na2S

Hope this helps.

4 0

2 years ago