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2 years ago

What element am I?

Chemistry

1 answer:

Serga [27]2 years ago

4 0

Answer:

H- hydrogen

Explanation: Hydrogen is in the first group meaning that it only has 1 valence electron and 1 energy level

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What is 31 degrees Celsius in Fahrenheit

vovangra [49]

87.8 , 31 Celsius=87.8(88) in Fahrenheit

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2 years ago

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A geneticist has discovered a new compound that is made up of a small number of identical

eduard

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They are monomers.

Explanation:

Edg 2020

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2 years ago

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Is condensation of ethanol a physical change or a chemical change

tigry1 [53]

Answer: Physical change

Explanation: Condensation is the phase change of a gas to the liquid state. There is no formation of new products so it does not undergoes chemical change. Any phase change is considered a physical change.

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2 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago

The PH of a 0.1 M MCl (M is an unknown cation) was found to be 4.7. Write the net ionic equation for the hydrolysis of M and its

Serggg [28]

Answer:

6.25 X10^{-9} = Ka

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$

Explanation:

The ionic equation for the hydrolysis of the cation of the given salt will be:

$M^{+} + H_{2}O ---> MOH + H^{+}$

The expression for Ka will be:

Ka = $\frac{[H^{+}][MOH]}{[M^{+}]}$

As given that the concentration of the salt is 0.1 M and pH of solution = 4.7, we can determine the concentration of Hydrogen ions from the pH

pH = -log [H⁺]

[H⁺] = antilog(-pH) = antilog (-4.7) = 2 X 10⁻⁵ M = [MOH]

Let us calculate Ka from this,

Ka = ${2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}$

The relation between Ka an Kb is

KaXKb =10⁻¹⁴

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$

6 0

2 years ago