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3 years ago

Maximum number of possible structural of C7H9N

Chemistry

1 answer:

MrRissso [65]3 years ago

3 0

There are 5 possible structural isomers with the molecular formula C₇H₉N.

Explanation:

You may find the chemical diagrams of the structural isomers in the attached picture.

Learn more about:

structural isomers

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Calcium and bromine have formed a bond. Leading up to this, calcium gave up electrons. It was a(n)

scZoUnD [109]

Leading up to this, calcium gave up 2 valence electrons and thus was denoted as a cation. These 2 electrons were transferred to bromine, which received an overall negative charge because of the addition of 2 valence electrons in its valence shell, and thus formed a negatively charged ion, an anion.

Both formed an ionic bond, due to the electrostatic charge of attraction between the 2 oppositely charged ions. If many ions of Ca and Br are present and numerous ionic bonds have formed it will undergo an arrangement which is that of an ionic lattice, type of structure.

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3 years ago

What is used to classify galaxies?star typesestimated agecolorshape

lianna [129]

The correct answer among the choices given is the last option. Galaxies are classified according to their shapes or visual morphology. There three main types of galaxies currently. They are the elliptical, spiral and irregular. Elliptical galaxies are like a spheriod or an elongated sphere. Spiral galaxies are characterized and distinguished by having a bulge, a disk and a halo. Lastly, irregular galaxies do not have regular or symmetrical shape.

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4 years ago

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

b) Br2 is the limiting reactant.

c) There will be formed 86.3 grams of BrF3

d) There will remain 0.326 moles of F2 = 12.97 grams

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

Br2 is the limiting reactant. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = 86.3 grams

7 0

3 years ago

In what form metals are found in nature

Alika [10]

Answer:

Metals in nature are found in rock form, mineral forms etc..

7 0

3 years ago

Read 2 more answers

A solution was made by adding water to 0.250 g of H2Z until the volume totaled 25.00 mL. Subsequent titration required 40.50 mL

Brilliant_brown [7]

Answer:

123 g/mol

Explanation:

Let's consider the neutralization reaction.

H₂Z(aq) + 2 NaOH(aq) → Na₂Z(aq) + 2 H₂O(ℓ)

The moles of NaOH that reacted are:

$40.50 \times 10^{-3} L \times \frac{0.100molNaOH}{1L} =4.05\times 10^{-3} molNaOH$

The molar ratio of H₂Z to NaOH is 1:2. The moles of H₂Z that reacted are:

$4.05\times 10^{-3} molNaOH \times \frac{1molH_{2}Z}{2molNaOH} =2.03\times 10^{-3} molH_{2}Z$

The molar mass of H₂Z is:

$\frac{0.250gH_{2}Z}{2.03\times 10^{-3} molH_{2}Z} =123g/mol$

7 0

3 years ago