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3 years ago

Match each alkane names and structures.

Chemistry

1 answer:

Scilla [17]3 years ago

6 0

First structure is CH₃-CH₃. It is an alkane with two carbon atoms hence it is called ethane.

Second structure is CH₃-CH₂-CH₂-CH₃. It is an alkane with four carbon atoms hence it is called butane.

Third structure is CH₄. It is an alkane with one carbon atom hence it is called methane.

Fourth structure is CH₃-CH₂-CH₃. It is an alkane with three carbon atoms hence it is called propane.

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What was done to the linear parent function, f(x) = x, to get the function g(x) = 1/6 x

s2008m [1.1K]

Answer:

Vertically Shrunk by a factor of 1/6

Explanation:

Parent Formula: f(x) = a(bx - c) + d

a - vertical shrink/stretch and x-reflections

b - horizontal shrink/stretch and y-reflections

c - horizontal movement left/right

d - vertical movement up/down

Since we are only modifying a, we are dealing with vertical shrink/stretch:

Since a < 1 (1/6 < 1), we are dealing with a vertical shrink of 1/6.

Since a > 0 (1/6 > 0), we do not have a reflection over the x-axis.

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3 years ago

How is molarity measured

Anuta_ua [19.1K]

Molarity (m) is defined as the number of moles to solute (n) the volume (v) of the solution in liters is important to note that the molarity is defined as moles of solute per liter of solution not moles of solute per liter of solute.

5 0

3 years ago

Read 2 more answers

The rate constant k of the second-order reaction CH3CHO→CH4+CO is 6.73×10−3Lmol s. The concentration of CH3CHO at t=50.0 seconds

Tanzania [10]

Answer:

The initial concentration of ethanal was 0.1590 mol/L.

Explanation:

Integrated rate law for second order kinetic:

$k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})$

k = Rate constant = $6.73\times 10^{-3} L mol s$

t = Time elapsed = 50.0 s

$[A]_o$ =initial concentration of ethanal

[A] = Concentration of ethanal left after time t = 0.151 mol/L

On substituting the value:

$6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})$

$[A]_o=0.1590 mol/L$

The initial concentration of ethanal was 0.1590 mol/L.

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2 years ago

Which two compounds are used in batteries? sulfuric acid and potassium hydroxide, lithium hydroxide and nitric acid, potassium h

AnnZ [28]

Answer:

sulphuric acid and ammonium hydroxide

Explanation:

They are different types of batteries. there are zinc carbon batteries, alkaline batteries, lithium carbon batteries, lead-acid batteries.

lithium is used for high performance devices such as smartphones, digital cameras and even electric cars.A variety of substances are used in lithium batteries, but a common combination is a lithium cobalt oxide cathode and a carbon anode

lead-acid batteries: This is the chemistry used in a typical car battery. The electrodes are usually made of lead dioxide and metallic lead, while the electrolyte is a sulfuric acid solution.

zinc carbon batteries; The anode is zinc, the cathode is manganese dioxide, and the electrolyte is ammonium chloride or zinc chloride.

alkaline batteries;The cathode is composed of a manganese dioxide mixture, while the anode is a zinc powder. It gets its name from the potassium hydroxide electrolyte, which is an alkaline substance.

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3 years ago

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The percent composition of all elements in HClO3 is:

ser-zykov [4K]

The answer is b. 1% H, 42% Cl, 57% O.

Let's first calculate a molar mass (M) of HClO₃. Molar mass is a mass of 1 mole of a substance. It can be expressed as the sum of relative atomic masses (Ar), which are masses of atoms of the elements:
M(HClO₃) = Ar(H) + Ar(Cl) + 3Ar(O)

Ar(H) = 1 u
Ar(Cl) = 35.5 u
Ar(O) = 16 u
M(HClO₃) = Ar(H) + Ar(Cl) + 3Ar(O) = 1 + 35.5 + 3 · 16 = 84.5 u

Now, we just need to calculate mass percent of each element using the proportion:
Ar : x = M : 100% ⇒ x = Ar ÷ M · 100%

Hydrogen: x = Ar(H) ÷ M · 100% = 1 ÷ 84.5 · 100% = 1.2% ≈ 1%
Chlorine: x = Ar(Cl) ÷ M · 100% = 35.5 ÷ 84.5 · 100% = 42%
Oxygen: x = 3Ar(O) ÷ M · 100% = 48 ÷ 84.5 · 100% = 56.8% ≈ 57%

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3 years ago