CHEMISTRY HIGH SCHOOL

All compounds containing carbon are organic.
a. True
b. False

Answers

Answer 1
Answer: False, not all compounds containing carbon is organic rather all organic compounds contain carbon and hydrogen. Some compounds containing carbon are considered to be inorganic like carbon dioxide, carbon tetrachloride and sodium bicarbonate.

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Balance the following oxidation-reduction reaction and indicate which atoms have undergone oxidation and reduction. ____HNO3 + ____S ---> ____NO + ____H2SO4

Answers

Explanation:

An equation that contains equal number of atoms on both reactant and product side is known as a balanced chemical equation.

Whereas when there will be increase in oxidation number of the atom then it is known as oxidation. When there is decrease in oxidation number of the atom then it is known as reduction.

For example, 4HNO_(3) + S \rightarrow NO + H_(2)SO_(4) + H_(2)O

Oxidation half reaction :S \rightarrow H_(2)SO_(4) + 6e^(-)

Reduction half reaction :HNO_(3) + 3e^(-) \rightarrow NO

Therefore, we can conclude that sulfur atom has undergone oxidation and nitrogen atom has undergone reduction.
The following oxidation-reduction reaction and indicate which atoms have undergone oxidation and reduction.
2HNO3 + S ---> 2NO + H2SO4

What are ionic compounds typically composed of?a metal anion and a nonmetal cation

two metal anions

a metal cation and a nonmetal anion

two nonmetal cations

Answers

ionic compounds are held together by ionic bonds are formed by the transfer of electrons from a metal to a non metal

Ionic bond is formed by the electrostatic attraction between the oppositely charged ions.

metals give out electrons and becomes positively charged. positively charged ions are called cations. so these are called metal cations

the electrons given out are taken in by non metals and become negatively charged. negatively charged ions are called anions. then they are called non metal anions.

so ionic compounds are formed by the electrostatic attraction between metal cations and non metal anions.

therefore ionic compounds are composed of a metal cation and a nonmetal anion
Ionic compounds are typically composed of a metal cation and nonmetal anion.

A vineyard has 145 acres of Chardonnay grapes. A particular soil supplement requires 5.50 g for every square meter of vineyard. How many kilograms of the soil supplement are required for the entire vineyard? (Recall that 1 km2 = 247 acres.) Express your answer in kilograms to three significant figures.

Answers

1.36.10 ^ 6 kg of supplements are required for the entire vineyard

Further explanation

7 main quantities have been determined based on international standards, namely:

  • 1. Length, meters (m)
  • 2. Time, second (s)
  • 3. Mass, kilograms (kg)
  • 4. Temperature, kelvin (K)
  • 5. Light intensity, candela (cd)
  • 6. Electric current, ampere (A)
  • 7. Amount of substance, mol (m)

Derivative magnitude is a quantity derived from one or more principal quantities. So in addition to 7 principal quantities, other quantities are derived quantities

An area with the formula length x width is a unit derived from the length of the principal. The international standard unit is square meters (m²).

other area units: Km², hm², dam², m², dm², cm², and mm²

Hectare is an SI unit

1 Hectare is equal to 100 a (Are) or 10000 m² (ten thousand square meters) or 100 x 100 m 1 Hectares = 2.47 acres

In the question , there is a 145 acres vineyard, with the supplement is given 5.5 gram / m² vineyard.

So for 1 km², a supplement  =

1 km² = 10⁶ m²

5.5 gram / m² = 5.5.10⁶ grams / km²

whereas 1 km² = 247 acres and vineyard = 145 acres so

5.5.10^6*(145)/(247)

= 3.2287.10⁶ grams

= 3.23.10³ kg (3 significant numbers: 3,2 and 3)

Learn more

Convert the following metric units of weight

brainly.com/question/11300981

conversion factor

brainly.com/question/2375946

convert a mass of 2.93 pounds to ounces

brainly.com/question/1386337

Keywords: area, convert, acres

The soil supplement required for the entire vineyard is \boxed{3.23 * {{10}^3}{\text{ kg}}}.

Further Explanation:

There are two types of units. One is basic or fundamental while the other ones are derived units. Basic units cannot be further reduced and other quantities are expressed in these units. Derived units are those that can need basic units to express themselves. Area, density, volume and velocity are some examples of derived units.

Seven basic units are present in the SI system. These are as follows:

1. Meter (m)

2. Kilogram (kg)

3. Second (s)

4. Kelvin (K)

5. Ampere (A)

6. Mole (mol)

7. Candela (Cd)

Firstly, the area of vineyard has to be converted into {\text{K}}{{\text{m}}^2}. The conversion factor for this is,

1{\text{ acre}} = \left( {\frac{1}{{247}}} \right){\text{ K}}{{\text{m}}^2}

Therefore the area of vineyard can be calculated as follows:

 \begin{aligned}{\text{Area of vineyard}} &= \left( {145{\text{ acres}}} \right)\left( {\frac{{1/247{\text{ K}}{{\text{m}}^2}}}{{1{\text{ acre}}}}} \right)\n&= 0.587045{\text{ K}}{{\text{m}}^2}\n\end{aligned}  

The area is again converted into {{\text{m}}^{\text{2}}}. The conversion factor for this is,

 1{\text{ K}}{{\text{m}}^2} = {10^6}{\text{ }}{{\text{m}}^2}

So the area of vineyard can be calculated as follows:

\begin{aligned}{\text{Area of vineyard}}&= \left( {0.587045{\text{ K}}{{\text{m}}^2}}\right)\left( {\frac{{{{10}^6}{\text{ }}{{\text{m}}^2}}}{{1{\text{ K}}{{\text{m}}^2}}}}\right)\n&= 587045{\text{ }}{{\text{m}}^2}\n\end{aligned}  

The amount of supplement required for the entire vineyard can be calculated as follows:

 \begin{aligned}{\text{Amount of supplement required}}&= \left( {587045{\text{ }}{{\text{m}}^2}} \right)\left( {\frac{{5.50{\text{ g}}}}{{1{\text{ }}{{\text{m}}^2}}}} \right)\n&=3.2287475 * {10^6}{\text{ g}}\n\end{aligned}

The amount of supplement is to be converted into kg. The conversion factor for this is,

 1{\text{ g}} = {\text{1}}{{\text{0}}^( - 3)}{\text{ kg}}

Therefore the amount of supplement can be calculated as follows:

 \begin{aligned}{\text{Amount of supplement}}&= \left({3.2287475 * {{10}^6}{\text{ g}}} \right)\left( {\frac{{{{10}^( - 3)}{\text{ kg}}}}{{1{\text{ g}}}}} \right)\n&= 3.2287475 * {10^3}{\text{ kg}}\n&\approx 3.23 * {10^3}{\text{ kg}}\n\end{aligned}

Learn more:

  1. What is the mass of 1 mole of viruses: brainly.com/question/8353774
  2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Basic concepts of chemistry

Keywords: supplement, vineyard, 5.50 g, 3.23*10^3 kg, conversion factor, basic units, fundamental units, derived units, area, volume, density, kg, m, Km, acre.

The atomic number of an atom is alwaysequal to the number of its. (1) protons, only. (2) neutrons, only. (3) protons plus neutrons. (4) protons plus electrons

Answers

You are correct, the atomic number is always equal to the number of protons in the element. It is the relative atomic mass that is equal to the number of protons+neutrons. Electrons have no mass

How do you calculate the number of moles when given the number of particles?

Answers

Answer:

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Convert 1.50μm2 to square meters

Answers

1 micro meter = 10^(-6) meters
1 μm ^2 = 1 μm*1μm = 10^(-6) *10^(-6) = 10^(-12) meters

1.5 μm^2 = 1.5 * 10^(-12) meters

μm^2 is a unit for surface. First you want to convert μm to meters which is unit for length and if you multiply units for length you get unit for surface.
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