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3 years ago

What was Johann Dobereiner’s contribution to the development of the periodic table?

Chemistry

2 answers:

Novay_Z [31]3 years ago

5 0

He identified triads of elements that had similar properties

umka21 [38]3 years ago

3 0

he introduced his law of triads. each triad was a group of three elements

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A sulfur atom weighs twice as much as an oxygen atom, If a chemist has 12grams of oxygen atoms, how many grams of sulfur would c

Arisa [49]

Answer: 24

Explanation: multiply 12 with 2

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3 years ago

How do we use biomass as an energy source?

kherson [118]

Biomass contains stored energy. That's because plants absorb energy from the sun through the process of photosynthesis. When biomass is burned, this storedenergy is released as heat. Burning biomass releases carbon dioxide

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4 years ago

What halogen can react with sodium chloride solution to produce chlorine?

ozzi

Answer:

As electro-negativity decreases from Florine to downwards in the group and only Florine is above Chlorine, so Florine should react with sodium chloride solution to produce chlorine.

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3 years ago

Please help me answer this question but in your own words!!!!

stiv31 [10]

Answer:

when sodium reacts with chlorine atoms, sodium will give an electron which is negative charged to chlorine.it makes sodium(+) and chlorine (-)..then sodium will attract chlorine to form sodium chloride(NaCl)

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3 years ago

A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con

Nataly_w [17]

Answer:

$\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

$Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}$

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

$Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J$

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

$n=1.400g*\frac{1mol}{89.05g} =0.01572mol$

Thus, we obtain:

$\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}$

Best regards!

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3 years ago