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2 years ago

QUICK!! WILL MARK BRAINLIEST!! what is the volume of 2.7 moles of co2 ?

Chemistry

1 answer:

Kay [80]2 years ago

3 0

Answer:

60. L CO₂

Explanation:

Step 1: Define

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm

Step 2: Use Dimensional Analysis (Assuming we are dealing with STP)

$2.7 \hspace{3} mol \hspace{3} CO_2(\frac{22.4 \hspace{3} L \hspace{3} CO_2}{1 \hspace{3} mol \hspace{3} CO_2} )$ = 60.48 L CO₂

Step 3: Simplify

We are given 2 sig figs.

60.48 L CO₂ ≈ 60. L CO₂

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Which of the following statements correctly distinguishes between an acidic and a basic solution? . A. An acidic solution contai

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Hence, letter C is the correct answer.

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2 years ago

33. Why does Fluorine exert a stronger pull than Chlorine on the valence electrons of another

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Answer:

Fluorine is much more reactive than chlorine (despite the lower electron affinity) because the energy released in other steps in its reactions more than makes up for the lower amount of energy released as electron affinity.

Explanation:

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2 years ago

How many electrons are shared in a double covalent bond?

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In a double covalent bonds there are 4 valence electrons that are shared.

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2 years ago

Read 2 more answers

Determine what is missing from this neutralization reaction: AgNO3+KCl→AgCl+−−−−

lapo4ka [179]

Answer:

The answer to your question is: KNO₃

Explanation:

AgNO3 + KCl → AgCl + −−−−

A. KNO3 this option is correct because it is a double replacement reaction then potassium must attached to NO₃.

B. KOH this product is not possible because there is no water to form OH⁻ ions.

C. Ag2K this product is not possible because both Ag and K are metals, then it is difficult that they attach.

D. KN2O This product is imposible to form, this option is wrong.

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2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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2 years ago