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3 years ago

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A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

2 answers:

andreev551 [17]3 years ago

7 0

Answer: The frequency of the wave is 0.5 hertz.

Explanation:

$\nu =\frac{c}{\lambda }$

$\nu$ = Frequency of the wave

$c$ = speed of the light in m/s

$\lambda$ = Wavelength of the wave.

Here in question we are given with speed of the infrared light. So, we will replace the value of speed of light(c) from the given value of the speed of the infrared light.

Speed of infrared light = 6 m/s

$\nu =\frac{\text{speed of infrared light}}{\lambda }=\frac{6 m/s}{12 m}=0.5 hertz$

The frequency of the wave is 0.5 hertz.

____ [38]3 years ago

4 0

$frequency=\frac{speed}{wavelength}$

Given
$speed=6\ m/s$
and
$wavelength=12\ m$
we get:
$frequency=\frac{6\ m/s}{12\ m}$
$\rightarrow frequency=0.5\ Hz$

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In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

gladu [14]

Answer:

$\large \boxed{\text{183 g}}$

Explanation:

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}$

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3 years ago

The pH of a basic solution is 9.77. What is [OH⁻]?

uysha [10]

The [OH⁻] of the solution is 5.37×10⁵ M.

<h3 /><h3>What is pOH?</h3>

This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.

To calculate the hydroxy ion [OH⁻] concentration we use the formula below.

Note:

pH+pOH = 14

pOH = 14-pH
pOH = 14-9.77
pOH = 4.27

Formula:

[OH⁻] = 1/ $10^{pOH}$ ................. Equation 1

Given:

pOH = 4.27

Substitute the value into equation 1

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Hence, The [OH⁻] of the solution is 5.37×10⁵ M.

Learn more about hydroxy ion concentration here: brainly.com/question/17090407

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2 years ago

Read 2 more answers

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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