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3 years ago

15

a gas sample is at 10 degress celcius if the pressure remains constant the volmue will increase when the temperature is changed

to

1 answer:

Sholpan [36]3 years ago

6 0

Answer:

we love points

Explanation:

we always will love points

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For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur

MakcuM [25]

Answer:

Moles of NO₂ = 0.158

Explanation:

SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )

According to the law of mass equation

$K_{c}$ = $\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}$

⇒ 3.10 = $\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}$ At equilibrium [SO₃] = [NO]

⇒ [NO₂] = $\frac{1}{6.3}$

⇒ [NO₂] = 0.158

So. number of moles of NO₂ at equilibrium added = 0.158

8 0

4 years ago

A tall pea plant (Tt) is crossbred with another tall pea plant (Tt). The following Punnett square shows the separated alleles fo

Luba_88 [7]

Answer:

B.

Explanation:

when you use punnett squares, you cross the different genes with themselves. if T is in the first box on both sides, the genome will be TT. if the box lines up to tt, it will be tt. the boxes on the left use the genes' first letter and the boxes on the right use the second letter.

7 0

3 years ago

Read 2 more answers

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

A gas bubble has a volume of 0.650 mL at the bottom of a lake, where the pressure is

astraxan [27]

Considering the Boyle's law, as the pressure decreases, volume increases and has a value of 2.246 mL.

<h3>Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

This law says that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

If an initial state 1 and a final state 2 are analyzed, Boyle's law is expressed as:

P1×V1=P2×V2

<h3>Volume at the surface of the lake</h3>

In this case, you know:

P1= 3.46 atm
V1= 0.650 mL
P2= 1 atm
V2= ?

Replacing in Boyle's law:

3.46 atm× 0.650 mL= 1 atm×V2

Solving:

V2= (3.46 atm× 0.650 mL)÷ 1 atm

<u><em>V2= 2.246 mL</em></u>

Finally, as the pressure decreases, volume increases and has a value of 2.246 mL.

Learn more about Boyle's law:

brainly.com/question/4147359

#SPJ1

4 0

2 years ago

Effect of fossil fuel emission on oceans?

BlackZzzverrR [31]

Fossil fuel emissions can create acid rain, which in turn increases the pH of ocean water, harming the organisms in the ocean.<span />

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4 years ago