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3 years ago

Please help, only a couple of days left!!!

Chemistry

2 answers:

Zinaida [17]3 years ago

4 0

Bit bang has to do it with ut

s344n2d4d5 [400]3 years ago

3 0

Answer:

the heat from the universe during the big bang

a.k.a.

ITS A

Explanation:

the universe is still expanding, so what is gonna be used to create that new matter that's expanding the universe? energy and heat from the beginning of the universe! the universe started off as this point that was probably smaller than a quantum foam and infinite heat and started expanding from there

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Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to b

Alja [10]

Answer:

439.7nm

Explanation:

Energy of a quantum can be calculated using below formula

E=hv...........eqn(1)

But v=λ/ c .........eqn(2)

If we substitute eqn(2) into eqn(1) we have

E= hc/(λ)

Where E= energy

h= Plank's constant= 6.62607004 × 10-34 m2 kg / s

c= speed of light

c= 2.998 × 10^8 m/s

λ= wavelength= ?

But the energy was given in Kj , it must be converted to Kj/ photon for unit consistency.

Energy E= 272 kJ/mol × 1mol/6.02× 10^23

Energy= 451.83× 10^-24 Kj/ photon

E= hc/(λ)...........eqn(1)

If we make λ subject of the formula

λ= hc/E

Then substitute the values we have

λ= [(6.626 × 10^-34) × (2.998 × 10^8)]/451.83× 10^-24

λ=(0.00043965) × (1Kj/1000J) × (10^9nm/1m)

λ=439.7nm

Hence, the longest wavelength of radiation with enough energy to break carbon-sulfur bonds is 439.7nm

4 0

3 years ago

How many grams are in 1.23 x 1020 atoms of arsenic?

Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0

2 years ago

Which substance is an electrolyte? 1. O2 2. Xe 3. C3H8 4. KNO3

Troyanec [42]

KNO3 is an electrolyte

3 0

3 years ago

Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2

Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

Molar mass of AgNO3 = 169.87 g/mol
Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0

2 years ago

The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$ .

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$ . In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$ . 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$ .

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$ . How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.

6 0

3 years ago