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2 years ago

CH, + , 0, → _CO, + Н,0 Balance chemical equation

Chemistry

1 answer:

Flura [38]2 years ago

6 0

Answer:

2 CH2 + 3 O2 = 2 CO2 + 2 H2O

Explanation:

This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.

Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.

When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.

However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.

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Properties of elements within a blank on the periodic table change in a predicatable way from one side of the table to the other

Vitek1552 [10]

Answer: oxygen

Explanation:

7 0

2 years ago

Metals and nonmetals all lesson explanation for class 8

WINSTONCH [101]

Answer:

Metalloids elements whose properties are intermediate between those of metals and solid nonmetals or semiconductors.

Some examples:

Chemical element.

Boron.

Semiconductor.

Arsenic.

Silicon.

Selenium.

Antimony.

Germanium.

7 0

2 years ago

Average Molarity for HCl is .391

Ira Lisetskai [31]

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>

8 0

2 years ago

Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

tatiyna

Answer:

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%
The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

4 0

2 years ago

Matt did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers la

Black_prince [1.1K]

Part 1 : Answer is only B substance is soluble in water.

In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.

Part 2 : Substance A is not soluble in water and substance B is soluble in water.

According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.

But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.

4 0

2 years ago

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