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2 years ago

On Darwin's tree of life, organisms at the base of

Chemistry

2 answers:

bazaltina [42]2 years ago

7 0

The correct answer is part A

masya89 [10]2 years ago

3 0

B - Evolve into the closest species on the tree’s trunk

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Calculate the volume of 12.0 g of helium at 100°c and 1.2 atm

hodyreva [135]

Volume = nRT/P

n = number of particles (moles)

R = universal gas constant (0.0821)

T = temperature (Kelvin)

P = pressure (atm)

(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He

Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K

Now we can set up the equation for volume: (3mol)(0.0821)(373.15)/1.2atm = 76.6 L of Helium gas

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2 years ago

How far above earth is the international space station

Rasek [7]

Answer:

248 miles

Explanation:

At an average altitude of 248 miles (400 kilometers) above Earth, the space station is the third brightest object in the sky.

4 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

What is the chemical composition of hot chocolate?

iris [78.8K]

Answer:

Hot chocolate is as straightforward as drinks go: at its core, it's milk, cocoa powder, and sugar. Despite its simplicity, this cold-weather classic is swirling with science. The backbone of any decent hot chocolate is milk. Beyond water, milk is perhaps the most basic and familiar substance to humans.

7 0

2 years ago

An organic chemistry student was studying the solubility of Methyl-N-acetyl-α-D-glucosaminide (1-O-methyl-GlcNAc), a derivative

strojnjashka [21]

Complete Question

The diagram for this question is shown on the second uploaded image

Answer:

The organic product obtained is shown on the first uploaded image

Explanation:

The process that lead to this product formation is known as oxidative cleavage which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them

4 0

2 years ago