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3 years ago

Pilots often use waves of about 2.340m to communicate. what is the frequency of the wave ?why?

Chemistry

1 answer:

uranmaximum [27]3 years ago

3 0

All of the various light waves move with the same speed, a value

<span>abbreviated (c) equal to 3.00 × 108 m/s. The frequency ([1]) that light waves pass a given point is</span>

measured in waves/second or simply ‘per second’ (1/s).

the frequency of the wave f = (3.00 × 10^8 m/s)/ 2.340 m

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Answer:

Explanation:

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3 years ago

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of c

Flauer [41]

Answer:

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

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Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

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6 0

3 years ago

Can someone please answer this?

Amanda [17]

Answer:-

0.229 L

Explanation:-

Molar mass of AgBr = 107.87 x 1 + 79.9 x 1

=187.77 grams mol-1

Mass of AgBr = 150 grams

Number of moles of AgBr = 150 grams / 187.77 gram mol-1

= 0.8 mol

The balanced chemical equation is

NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)

From the equation we can see that

1 mol of AgBr is produced from 1 mol of AgNO3.

∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.

Strength of AgNO3 = 3.5 M

Volume of AgNO3 required = Number of moles / strength

= 0.8 moles / 3.5

=0.229 L

8 0

3 years ago

What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M

Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0

3 years ago