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Naily [24]
2 years ago
13

Which of the following is NOT useful in

Chemistry
1 answer:
Kipish [7]2 years ago
4 0
The price would not be a physical property
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What is Chemistry and how do you relate in real life?
Neko [114]
It is known that chemistry is a BIG part of our everyday lives. You can find chemistry in daily life in foods you eat, air you breathe, soap, your emotions and literally every object you can see or touch. For example, Chemistry explains how food changes as you cook it, how it rots, how to preserve food, how your body uses the food you eat, and how ingredients interact to make food.
Hope it helps! :)
3 0
3 years ago
Read 2 more answers
Is that the right answer
julsineya [31]
Is what the right answer
8 0
3 years ago
When butter is heated it melts and when that melted butter cools and solidifies the process called
avanturin [10]
<span>I'm pretty sure it is called condensation</span>
3 0
3 years ago
What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?
alina1380 [7]

Mass of H₂ needed to react with O₂ : 1.092 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Reaction

O₂(g) + 2H₂(g) → 2H₂O(g)

mass of O₂ : 8.75 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{8.75}{32}=0.273

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

\tt \dfrac{2}{1}\times 0.273=0.546

Mass H₂ (MW=2 g/mol) :

\tt 0.546\times 2=1.092~g

4 0
3 years ago
Read 2 more answers
What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?
Varvara68 [4.7K]

The dissociation of formic acid is:

HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

k_a = \frac{[HCOO^{-}]  [H^{+}]}{HCOOH}

Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

pH = 2.75

pH = -log[H^{+}]

[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}

pk_a = 3.75

k_a = 10^{-3.75} = 1.78\times 10^{-4}

Substituting the values in the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}

Hence, the ratio is \frac{1}{10}.

4 0
3 years ago
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