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sergij07 [2.7K]
2 years ago
12

What color will a white dress appear if white light is passed through a red filter. Explain.

Chemistry
1 answer:
MatroZZZ [7]2 years ago
5 0

"If you pass white light through a red filter, then red light comes out the other side. This is because the red filter only allows red light through. The other colors (wavelengths) of the spectrum are absorbed. Similarly, a green filter only allows green light through."

I'm not too sure it either wouldn't be too visible or it would have a darker color..lmk if you have questions

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(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs 15.0 g of ethanol fo
o-na [289]

<u>Answer:</u>

<u>For a:</u> The density of the sample of copper is 8.96g/cm^3

<u>For b:</u> The volume of ethanol needed is 19.0 mL

<u>For c:</u> The mass of mercury is 340. grams

<u>Explanation:</u>

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}      ......(1)

  • <u>For a:</u>

Mass of copper = 374.5 g

Volume of copper = 41.8cm^3

Putting values in equation 1, we get:

\text{Density of copper}=\frac{374.5g}{41.8cm^3}\\\\\text{Density of copper}=8.96g/cm^3

Hence, the density of the sample of copper is 8.96g/cm^3

  • <u>For b:</u>

Mass of ethanol = 15.0 g

Density of ethanol = 0.789 g/mL

Putting values in equation 1, we get:

0.789g/mL=\frac{15.0g}{\text{Volume of ethanol}}\\\\\text{Volume of ethanol}=\frac{15.0g}{0.789g/mL}=19.0mL

Hence, the volume of ethanol needed is 19.0 mL

  • <u>For c:</u>

Volume of mercury = 25.0 mL

Density of mercury = 13.6 g/mL

Putting values in equation 1, we get:

13.6g/mL=\frac{\text{Mass of mercury}}{25.0mL}\\\\\text{Mass of mercury}=(13.6g/mL\times 25.0mL)=340.g

Hence, the mass of mercury is 340. grams

5 0
3 years ago
The final marks in a statistics course are normally distributed with a mean of 70 and a standard deviation of 10. The professor
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Use inversenormal on a calculator and type in .1 , 70 , 10 . That percent is for A and will determine at what mark an A will be. Do the same for the rest of the grades but change the first argument in the calculation
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Conservation of ____________ energy is only valid when it occurs in a situation with no frictional force.
Flura [38]
Potential energy......
5 0
3 years ago
With a well labeled diagram explain the stages of meiosis and mitosis​
NNADVOKAT [17]

Answer:

hope that helps

7 0
3 years ago
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thi
77julia77 [94]

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]

Using D_o and E_a values obtained from the graph:

Thus;

\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr

B= 386 \  exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\  = 0.1719 \ \mu m^2/hr

So, the initial time required to grow oxidation is expressed as:

t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)

where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\  B = 0.1719

∴

2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)

2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2  \\ \\ t_o(initial) = 1.4267 \ hr

NOW;

1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\  1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453

d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}

d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}

d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}

d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \   \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}

d_o =0.02609 \ OR \   -0.0939

Thus; since we will consider the positive sign, the initial thickness d_o is ;

≅ 0.261 μm

3 0
3 years ago
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