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3 years ago

Substances which will glow in the dark after being exposed to sunlight are called:

Chemistry

2 answers:

DiKsa [7]3 years ago

7 0

Answer:

phosphorescent substances.

Explanation:

nordsb [41]3 years ago

6 0

It is called phosphorescent substances.

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At 580 k, 6. 42% of the molecules are in the chair form. calculate the value of the equilibrium constant for the reaction as wri

Ann [662]

Equilibrium chemical reaction between chair and boat forms is:
Kc = $\frac{[Boat form]}{[Chair form]}$
Suppose the total number of molecules is x
Number of molecules of chair form is 6.42/100 x
Number of molecules of boat form is 93.58/100 x
Kc = $\frac{(0.9358x)}{(0.0642x)}$ = 14.6

5 0

3 years ago

The electronic configuration of an element is given below.

Amiraneli [1.4K]

The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is

it is reactive because it has to lose one electron to have a full outermost energy level.

Explanation

Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.

sodium has one electron in the outermost energy level.

for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)

4 0

3 years ago

Select multiple choices. Thank you

AnnZ [28]

D, systems of writing

4 0

3 years ago

Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the

Ganezh [65]

Answer:

57.48%

Explanation:

Calculate the mass of 1 mole of malachite:

MM Cu = 63.55

MM O = 16.00

MM H = 1.01

MM C = 12.01

$(Cu_{2}(OH)_{2}CO_{3})$

A mole of malachite has:

2 moles of Cu

5 moles of O

2 moles of H

1 mole of C

MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)

MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01

MW Malachite = 221.13

Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1

Now divide the mass of Cu by the mass of Malachite

$%Cu = \frac{127.1}{221.13} =0.5748=57.48%$

7 0

3 years ago

A certain compound was found to have the molecular formula C5H12O2. To which of the following compound classes could the compoun

IgorC [24]

Answer : The given compound belongs to ether and alcohol.

Explanation :

The chemical formula of the given compound is, $C_5H_{12}O_2$

First we have to calculate the degree of unsaturation.

Formula used:

Degree of unsaturation = $\frac{2C+2+N-X-H}{2}$

where,

C = number of carbon

H = number of hydrogen

N = number of nitrogen

X = number of halogen

Degree of unsaturation = $\frac{2\times 5+2+0-0-12}{2}=0$

The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.

Thus, the given compound belongs to ether and alcohol.

6 0

3 years ago