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4 years ago

some infrared radiation reaches earth and sum does not with part does reach earth longer wave length or short wave length

1 answer:

4 0

The answer is long wave length because long wave lengths contians less energy, and would not harm living things such as plants and animals. the more engey a wave length has, the less harmful it is.

short wave length: lots of energy, extremely hot. (examples: gamma rays, and UV (ultraviolet) rays.

long wave lengths: not much energy, safe for humans and other life on Earth.

hopefully this helps.

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When ions form they have the electron configuration of: A) Alkali Metals

cricket20 [7]

The answerrrrrr isssss B I think

3 0

3 years ago

Read 2 more answers

Calculate the molecular mass of N2O3

Fynjy0 [20]

the answer is 76.01g/mol

and the density:1.4g/cm³

5 0

3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is $1.3\times 10^2g$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

3 years ago

A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic ener

ICE Princess25 [194]

Explanation:

The kinetic energy of translation

$E_1=\frac{1}{2}mv^2$

m= mass v= linear velocity

The kinetic energy of rotation

$E_2=\frac{1}{2}I\omega^2$

I= MOI of the thin walled sphere =kmR^2

where ω= v/R= angular velocity

$E_2=\frac{1}{2}kmR^2\frac{v}{R}^2$

Then

$\frac{E_1}{E_2} = \frac{0.5mv^2}{0.5kmR^2\frac{v}{R}^2 }$

=1/k

solid sphere: k=0.4; E1/E2 =1/0.4 = 2.5;

hollow sphere: k=2/3; E1/E2 = 1.5

3 0

3 years ago

How many atoms are in 52.3 g of lithium hypochlorite (LiClO)?

garri49 [273]

Answer:

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

Explanation:

To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.

52.3 g ÷ 58.4 g/mol = 0.896 mol

Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.

0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.

Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.

5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

7 0

3 years ago