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2 years ago

All BUT one is a quantitative observation. 420 dogs .667 cm big grasshoppers

Chemistry

1 answer:

RSB [31]2 years ago

7 0

Big grasshoppers, because it does not involve a number.

Hope this helps! <3

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Doping Se with P would produce a(n) ________ semiconductor with ____________ conductivity compared to pure Se.

deff fn [24]

Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se

the reason is P dopant will introduce holes in the Se as P has lesser valence electron

4 0

2 years ago

Which of the following best describes technology

kotegsom [21]

were are the choices

5 0

2 years ago

Read 2 more answers

An alloy of bronze is manufactured by melting 51.2 g of copper with 6.84 g of tin. What is the percent copper in the bronze?

Harlamova29_29 [7]

Answer:

Percentage of copper = 88%

Explanation:

Given data:

Mass of copper = 51.2 g

Mass of tin = 6.84 g

Percentage of copper = ?

Solution:

Formula:

Percentage of copper = mass of copper / total mass × 100

Now we will determine the total mass:

Total mass = mass of copper + mass of tin

Total mass = 51.2 g + 6.84 g

Total mass = 58.04 g

Now we will calculate the percentage of copper.

Percentage of copper = 51.2 g / 58.04 g × 100

Percentage of copper = 0.88 × 100

Percentage of copper = 88%

5 0

2 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago

True or false: covalent compounds are formed when two nonmetals share valance electrons

WITCHER [35]

Answer:

true

Explanation:

H2O : 2 covalent bonds are established between O atom and 2 H atom (H--O--H) by mutual shearing of their valence electrons.

8 0

2 years ago