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Aloiza [94]
3 years ago
11

Pls help and match them

Chemistry
1 answer:
slega [8]3 years ago
3 0

Answer:

Simile: C

Personification: D

Symbol: A

Hyperbole: B

Metaphor: E

Explanation:

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Answer questions a-c about the Bronsted acid-base reaction below using the identifying letters A-D below each structure. The pKa
GuDViN [60]

Answer:

a  H2CO3 b HCO3- and c H+ and HCO3-

Explanation:

As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.

  The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.

  The conjugate base of H2CO3 is bicarbonate ion(HCO3-)

c The species that predorminates at equilibrium are H+ and HCO3-

8 0
3 years ago
This is the equation for the combustion of propane.
yan [13]

Answer:

B

Explanation:

Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:

reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

8 0
3 years ago
Read 2 more answers
A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calc
disa [49]

Answer:

specific heat = 0.951 j/g·°C

Explanation:

Heat flow equation  => q = m·c·ΔT

q = heat flow = 4817 joules

m = mass in grams = 140 grams Aluminum

c = specific heat = ?

ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C

q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C

6 0
3 years ago
An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 113 amu and 115 amu, with natural abunda
DENIUS [597]

Answer: Its average atomic mass is 114.9 amu

Explanation:

Mass of isotope 1 = 113 amu

% abundance of isotope 1 = 5% = \frac{5}{100}=0.05

Mass of isotope 2 = 115 amu

% abundance of isotope 2 = 95% = \frac{95}{100}=0.95

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(113\times 0.05)+(115\times 0.95)]

A=114.9amu

Thus its average atomic mass is 114.9 amu

5 0
3 years ago
Forecasting the weather for a specific area is relatively easy because of all the technology that is available.
AleksAgata [21]

Answer:

A

Explanation:

4 0
3 years ago
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