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4 years ago

12

An ice cube with a volume of 45.0ml and a density of 0.9000g/cm3 floats in a liquid with a density of 1.36g/ml. what volume of t

he cube is submerged in the liquid

1 answer:

Grace [21]4 years ago

5 0

Answer : The volume of the cube submerged in the liquid is, 29.8 mL

Explanation :

First we have to determine the mass of ice.

Formula used :

$\text{Mass of ice}=\text{Density of ice}\times \text{Volume of ice}$

Given:

Density of ice = $0.9000g/cm^3=0.9000g/mL$

Volume of ice = 45.0 mL

$\text{Mass of ice}=0.9000g/mL\times 45.0mL$

$\text{Mass of ice}=40.5g$

The cube will float when 40.5 g of liquid is displaced.

Now we have to determine the volume of the cube is submerged in the liquid.

$\text{Volume of ice}=\frac{\text{Mass of liquid}}{\text{Density of liquid}}$

$\text{Volume of ice}=\frac{40.5g}{1.36g/mL}$

$\text{Volume of ice}=29.8mL$

Thus, the volume of the cube submerged in the liquid is, 29.8 mL

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CO2+2H2O=H3O^++HCO3^-

bekas [8.4K]

<u>Answer:</u>

<em>Bronsted Lowry Theory</em>

<em> $CO_2$ is the acid </em>

<em> $H_2 O$ is the base </em>

<em> $H_3 O^+$ is the conjugate acid and </em>

<em> $HCO_3^-$ is the conjugate base</em>

<u>Explanation:</u>

Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

A base is a substance which can accept one or more protons

Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.

Thus Hydrogen atom has no Charge and it is neutral.

When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

Thus Hydrogen ion has a positive charge.

Hydrogen ion is also called as a proton since it has only 1 proton in it.

Hydrogen ion in water that is,

$H^++H_2O\Rightarrow H_3O^+$

$H_3O^+$ is called as Hydronium ion.

Acid loses H+ (proton) to form conjugate Base

Base gains H+ (proton) to form conjugate Acid

For example

Let us consider the example given in the question

$CO_2+2H_2 O>H_3 O^+ +HCO_3^-$ can be written as (removed 1 $H_2 O$ from both the sides )

$CO_2+H_2 O>H^++HCO_3^-$ or

$CO_2+H_2 O > H_2 CO_3$ reversing the equation

$H_2 CO_3+H_2 O>H_3 O^+ +HCO_3^-$

$H_2 CO_3$ is the acid which donates $H^+$ to form $HCO_3^-$

$H_2 O$ is the base which gains $H^+$ to form $H_3 O^+$

$HCO_3^-$ is the conjugate base and $H_3 O^+$ is the conjugate acid

So

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

(Answer)

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Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

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$l = 1$ .
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Explanation:

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