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2 years ago

What superhero gained his powers by becoming a transgenic organism?

Chemistry

1 answer:

Karolina [17]2 years ago

7 0

Answer:

B.spiderman

Explanation:

DNA has been transferred to another species. Spiderman became Spiderman when DNA from a spider combined with a human.

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Asap... define pure substances...

gtnhenbr [62]

A pure substance refers to an element or a compound that has no component of another compound or element. Pure substances are made of only one type of atom or molecule. Hydrogen gas and pure iron are examples of pure substances. Hydrogen consists of hydrogen atoms only while iron consists of only iron atoms. Mixing two pure substances results in a mixture. To separate the two, scientists use a method known as filtration. Mixtures can either be homogeneous or heterogeneous. The measure used to determine how pure a substance may be called purity. Besides hydrogen and iron, other pure substances include gold, diamonds, sugar, and baking soda.

3 0

3 years ago

Read 2 more answers

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$

By knowing the moles of gold:

$n_{Au}=100g*\frac{1mol}{197g} =0.508mol$

It is possible to write the aforementioned formula in terms of the variable $x$ representing the moles of silver:

$20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]$

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

$x=n_{Ag}=21.0molAg$

So the mass is:

$m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g$

Best regards!

3 0

3 years ago

#1: The element that loses electrons in a chemical reaction is said to have been which of the following?

Mrrafil [7]

#1: The element that loses electrons in a chemical reaction is said to have been which of the following? A. reduced

B. oxidized

C. combusted

D. rusted

**idk, is it reduced or oxidized??

Answer: oxidized. The element that loses electrons increases its oxidation number (becomes more positve or less negative) and this is oxidation.

Which type of reaction occurs if one element gains electrons and another loses electrons?

A. combustion

B. synthesis

C. oxidation-reduction

D. double-displacement

**my answer: C

Yes, oxidation-reduction

#3: What is the oxidation number for phosphorus in H4P2O7?

A. +5

B. +3

C. +1

D. - 1

**my answer: A. +5

This is the way to calculate it 4(+1) + 2x + 7(-2) = 0 => x = +5

#4: What is the oxidation number for phosphorus in Na2HPO3?

A. +5

B. +3

C. +1

D. - 1

**my answer; B. +3

is that right??

Right

2(+1) + 1 + x + 3(-2) = 0 => x = 6 -1 -2 = 3 (positive)

5 0

3 years ago

Read 2 more answers

Plant and animal cells release energy stored in the bonds of glucose molecules when they perform _________. Question 1 options:

Dmitry_Shevchenko [17]

Cellular respiration

3 0

3 years ago

Read 2 more answers

Ammonia and oxygen react to form nitrogen and water.

Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0

3 years ago