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Gala2k [10]
2 years ago
11

What is necessary to produce visible light from chemical compounds?

Chemistry
1 answer:
erastova [34]2 years ago
5 0

Answer:

the molecule must contain either pi bonds or atoms with non-bonding orbitals.

Explanation:

A non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen.

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A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at
weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

-Q_1=Q_2

Mass of tin = m_1=18.5 g

Specific heat capacity of tin = c_1=0.21 J/g^oC

Initial temperature of the tin = T_1=97.38^oC

Final temperature = T_2=T

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.7 g

Specific heat capacity of water= c_2=4.184 J/g^oC

Initial temperature of the water = T_3=21.52^oC

Final temperature of water = T_2=T

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0
3 years ago
Question 6 of 10
Marta_Voda [28]

Answer:

B. A chemist

<em>Hope it's help :)</em>

5 0
3 years ago
What is the molar mass, in grams, of a mole of an element equal to?
prohojiy [21]
Technically molar mass cannot be in grams, it is in grams per mole. and it refers to a specific number of molecules of a substance, therefore substances have different molar masses because the elements have different weights. for example having 10 water molecules would be a lot heavier than having 10 air molecules
6 0
2 years ago
How many protons does gold have
shusha [124]

Answer:

gold has 79 protons

Explanation:

i looked it up lol

8 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
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