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3 years ago

6. What is the wavelength (in meters) of the electromagnetic carrier wave transmitted by The Sports

Chemistry

1 answer:

Vsevolod [243]3 years ago

8 0

Answer:

wavelength = 2.81 m

Explanation:

Given that,

The frequency of the Sports Fan radio station, f = 106.4 MHz

We need to find the wavelength (in meters) of the electromagnetic carrier wave. The relation between frequency and wavelength is given by :

$c=f\lambda$

Where,

$\lambda$ is wavelength

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{106.4\times 10^6}\\\\\lambda=2.81\ m$

So, the wavelength of the electromagnetic carrier wave is 2.81 m.

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How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO3 + CaC

nexus9112 [7]

Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with $\frac{6.2}{2}$ = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

6 0

3 years ago

The unique odors and flavors of many fruits are primarily due to small quantities of a certain class of organic compounds. The e

SpyIntel [72]

Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16

Hope this helps

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3 years ago

And non-flammable gases Noble gases are that have low chemical

gtnhenbr [62]

Answer:

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3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$