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umka2103 [35]
3 years ago
14

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

Chemistry
1 answer:
elixir [45]3 years ago
8 0

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

<u>Step 1: </u>Data given

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

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Which of the following solutions is more concentrated?<br> 0.50M KCl or 5.0% (w/v) KCl
HACTEHA [7]

.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.

6 0
3 years ago
Why sometimes the electrons of an atom transfer to higher energy level​
Elenna [48]

Answer:

According to Bohr, the amount of energy needed to move an electron from one zone to another is a fixed, finite amount. ... The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level. This situation is unstable, however.

8 0
3 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
3 years ago
A 6.165 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.27 grams of CO2 and 3
e-lub [12.9K]

Answer:

Explanation:

mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g

mass of hydrogen ( H ) in 3.363 g of H₂O = 2  x 3.363 / 18

= .373 g

These masses would have come from the sample of 6.165 g .

Rest of 6.165 g of sample is oxygen .

So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g

Ratio of C  , H , O in the sample

2.8 : .373 : 2.992

C: H : O : : 2.8 : .373 : 2.992

Ratio of moles

C: H : O : : 2.8/12 : .373/1 : 2.992 / 16

C: H : O : : .2333 : .373 : .187

C: H : O : : .2333/.187 : .373/.187 : .187/.187

C: H : O : : 1.247 : 1.99 : 1

C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )

Hence empirical formula

C₅H₈O₄

Molecular formula ( C₅H₈O₄ )n

n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132

n x ( 60 + 8 + 64 ) = 132

n = 1

Molecular formula = C₅H₈O₄.

3 0
3 years ago
How are animal-like and plant-like protists are similar and different?​
IrinaVladis [17]

Answer:

The main way animal-like protists differ from plant-like protists is in the way they get energy. Animal-like protists are heterotrophs. They have to eat other protists or bacteria in order to get energy, much as animals have to eat plants or other animals to get their energy. Plant-like protists, on the other hand, are autotrophs. They can make their own energy from the sun or other sources just as plants can.

Explanation:

hope that helps ☺️

4 0
3 years ago
Read 2 more answers
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