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2 years ago

What is the layer of the earths atmosphere that we live in

Chemistry

2 answers:

FrozenT [24]2 years ago

8 0

Answer:

troposphere

Explanation:

The troposphere is the lowest layer of the atmosphere. This is the layer where we live and where weather happens. Temperature in this layer generally decreases with height. The boundary between the stratosphere and the troposphere is called the tropopause.

Maurinko [17]2 years ago

3 0

The layer that we live in is the Troposphere.
Hope this answer is correct.

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A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

OLga [1]

The molar mass of Naphthalene is 128g/mol
Therefore; a mass of 1.64 g of Naphthalene contains'
= 1.64g/128 g
= 0.0128 moles
But, from the Avogadro's law 1 mole of a substance contains 6.022 × 10^23 particles
Therefore 1 mole of Naphthalene contains 6.022×10^23 molecules
Hence; 0.0128 moles × 6.022 ×10^23 molecules
= 7.716 × 10^21 molecules

3 0

2 years ago

This is agree or disagree I need help ASAP!!

PIT_PIT [208]

All of them are agree :)

8 0

2 years ago

Read 2 more answers

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$