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Artemon [7]
3 years ago
12

a sample Of water is heated by 520 cal in the temperature rises by 10 K what is the mass of the water sample in grams?

Chemistry
1 answer:
SVEN [57.7K]3 years ago
4 0
<h3>Answer:</h3>

52 g water

<h3>Explanation:</h3>

In this question we are given;

Amount of heat, Q = 520 calories

Change in temperature, K = 10 K

We are required to calculate the mass of water.

Amount of heat is calculated by multiplying mass by specific heat capacity by change in temperature.

Q = m × c × Δt

The specific heat capacity,c of water = 1 Cal/g°C or 4.184 J/g°C

Rearranging the formula;

m = Q ÷ cΔt

   = 520 Cal ÷ (1 Cal/g°C×10 K)

   = 52 g

Thus, the mass of water sample is 52 g

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Given the balanced equation representing a reaction: h2 → h h what occurs during this reaction? 1. energy is absorbed as bonds a
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I believe the correct answer from the choices listed above is option 2. From the balanced equation, the energy is absorbed as the bonds are broken. Energy is needed to be supplied for the bonds to be broken.

Hope this answers the question. Have a nice day.
8 0
3 years ago
Read 2 more answers
A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what
ludmilkaskok [199]

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

  • m = mass
  • c = specific heat capacity
  • ∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

8 0
2 years ago
"Nucleic acid was isolated by lysing cells in detergent, guanidine isothiocyanate, and DNAse, followed by extraction in an acid
Romashka [77]

Answer:

RNA

Explanation:

The nucleic isolated was quantified by 2 types of fluorescent dyes, which binds only to specific type of nucleic acid. You can find the information on the specificity of dyes on your labsheet or dye supplier's website. For our dyes, however:

  • SyBr Green II only emits fluorescent light when it binds to RNA or Single Strand DNA
  • PicoGreen only fluoresce when bound to Double Strand DNA

With these information, we know that our nucleic acid is either single strand DNA or RNA, however, it is also mentioned that

"Nucleic acid was isolated by lysing cells in detergent, guanidine isothiocyanate, and DNAse"

DNAse, which is an enzyme that breaks down DNA, destroys any extracted DNA during the process, thus, the only possible nucleic acid we have isolated is RNA

8 0
3 years ago
The current global population is 7.7 billion people, and the WHO estimates the minimum water need per person per day is 15 L. An
anygoal [31]

Answer:

a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population.  ( t = 5.489 E-13 day )

b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population

Explanation:

Let P = 7.7 billion people = 7.7 E12 person

∴ water needed for the total population for one day:

⇒ water amount  = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day

⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year

∴ freshwater available:

freshwater = 6.34 E2 Km² H2O

how long will this water sustain the current population?

⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day

this amount of freshwater does not even meet the need of the current global population.

∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature

⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )

⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year

⇒ Py > water amount

the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.

4 0
3 years ago
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