How much heat is liberated when 40 gram H, gas react with excess O2 gas

Name and explain the two types of atom bonding

nordsb [41]

Ionic bonding which is the attraction between the cations(metal) and anions (non metal).

Metallic bonding is the electrostatic force of attraction between the fixed cations and the delocalized electrons.

7 0

4 years ago

Worksheet Ch.3: % Composition, Molarity, and Other Units for

Ann [662]

The percentage composition by mass of the elements in the compound are:

Ai. The percentage composition by mass of Na in 3.65 g of NaF is 54.8%

Aii. The percentage composition by mass of F in 3.65 g of NaF is 45.2%

Bi. The percentage composition by mass of Zinc in the compound is 67.1%

Bii. The percentage composition by mass of Sulfur in the compound is 32.9%

Percentage composition by mass of an element in a compound can be obtained by using the following formula: $Percentage = \frac{mass of Element}{mass of compound} * 100$

Ai. Determination of the percentage of Na in 3.65 g of NaF

Mass of Na = 2 g

Mass of NaF = 3.65 g

<h3>Percentage of Na =? </h3>

$Percentage = \frac{mass of element }{mass of compound} * 100\\\\Percentage of Na = \frac{2}{3.65} * 100$

<h3>Percentage of Na = 54.8%</h3>

Aii. Determination of the percentage of F in 3.65 g of NaF

Mass of F = 1.65 g

Mass of NaF = 3.65 g

<h3>Percentage of F =? </h3>

$Percentage = \frac{mass of element }{mass of compound } * 100\\\\Percentage of F = \frac{1.65}{3.65} * 100\\\\$

<h3>Percentage of F = 45.2%</h3>

Bi. Determination of the percentage composition of Zn in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

<h3>Percentage of Zn =? </h3>

$Percentage = \frac{mass of element}{mass of compound} * 100\\\\Percentage of Zn = \frac{32.69}{48.72} *100\\\\$

<h3>Percentage of Zn = 67.1%</h3>

Bii. Determination of the percentage composition of Sulphur in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

Mass of S = 48.72 – 32.69 g

Mass of S = 16.03 g

<h3>Percentage of S =? </h3>

$Percentage = \frac{mass of element}{mass of compound} * 100\\\\Percentage of S = \frac{16.03}{48.72} * 100\\\\$

<h3>Percentage of S = 32.9%</h3>

Learn more: brainly.com/question/1350382

3 0

3 years ago

Fungi have eukaryotic cells (large cells with a nucleus), like animals and plants<br> True or false?

Serggg [28]

True your answer is true

8 0

3 years ago

1. What is the relationship between the concentration of the hydronium and hydroxide

Hitman42 [59]

Answer:

- Neutral solutions: concentration of hydronium equals the concentration of hydroxide.

- Acid solutions: concentration of hydronium is greater than the concentration of hydroxide.

- Basic solutions concentration of hydronium is lower than the concentration of hydroxide.

Explanation:

Hello,

It is widely known that the pH of water is 7, therefore the pOH of water is also 7 based on:

$pH+pOH=14$

In such a way, we can compute the concentration of hydronium and hydroxide ions as shown below:

$pH=-log([H^+])\\$

$[H^+]=10^{-pH}=10^{-7}=1x10^{-7}M$

$pOH=-log([OH^-])$

$[OH^-]=10^{-pOH}=10^{-7}=1x10^{-7}M$

Thus, we notice that the relationship between the concentration of the hydronium is equal for water or neutral solutions. Moreover, if we talk about acid solutions, pH<OH therefore the concentration of hydronium is greater than the concentration of hydroxide. On the other hand if we talk about basic solutions, pH>OH therefore the concentration of hydronium is lower than the concentration of hydroxide.

Best regards.

4 0

3 years ago

The equilibrium expression for kp for the reaction below is __________. n2 (g) o2 (g) 2no (g)

PolarNik [594]

For the reaction;
N2(g) + O2(g) = 2NO(g)

Kp = pNO²/ pN₂pO₂; (No units)
where;
pNO is the partial pressure of NO;
pN₂ is the partial pressure of nitrogen
pO₂ is the partial pressure of Oxygen
The equilibrium constant Kp is deduced from the balanced chemical equation for a reversible reaction, NOT experimental data as is the case for rate expressions in kinetics.
Kp changes with temperature considerably changing the position of an equilibrium, and, at a constant temperature, and therefore constant K, the position of an equilibrium can change significantly depending on relative concentrations/pressures of 'reactants' and 'products'.

8 0

3 years ago