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3 years ago

Evaluate the carbon dioxide molecule. Explain how to determine if double or triple bonds exist in the molecule.

Chemistry

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

It contain double Bond.

Explanation:

To determine weather the bond is double or triple simply check the electron involved in mutual sharing of an electron if 2 electron takes parts it said to be double or if 3 it said to be triple.

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Which of the following why use approximately represent the number of atoms in one mole of gold

natka813 [3]

Answer is: D. 6.02 x 1023.
Because this is Avogadro constant (the number of constituent particles, in this example atoms of gold that are contained in the amount of substance given by one mole). The mole is the unit of measurement for amount of substance, the mole is an SI base unit, with the unit symbol mol.

6 0

3 years ago

Read 2 more answers

What are the empirical formula and empirical formula mass for C10H30O10?

AysviL [449]

Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

C : H : O

10 : 30 : 10

Divide the ratio by two to get simplest ratio

C : H : O

10/10 : 30/10 : 10/10

1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

Empirical formula of C₁₀H₃₀O₁₀ = CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

4 0

3 years ago

An atomic number stands for the number of _____. neutrons in the nucleus of an atom protons in the nucleus of an atom valence el

Effectus [21]

Answer:

atoms or electrons

Explanation:

but l guess electrons is the best answer

5 0

3 years ago

Nts

a_sh-v [17]

The reaction is a synthesis reaction.

<h3>What are synthesis reactions?</h3>

Synthesis reactions are one of the numerous reactions in chemistry. These kinds of reactions involve the combination of two atoms of different elements resulting in the formation of new compounds.

Synthesis reactions are sometimes referred to as combination reactions, simply because they have to do with the combination of two or more atoms of different elements into a single compound.

The different elements that combine are referred to as the reactants while the result of the combination is known as the product.

For example: A + B --> AB

In the illustrated reaction, N2 combined with H2 to form NH3. N2 and H2 gases are the reactants while NH3 is the only product formed from the two reactants.

This is a good example of a synthesis/combination reaction.

More on synthesis reactions can be found here: brainly.com/question/24936069

#SPJ1

6 0

1 year ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

3 years ago