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WARRIOR [948]
3 years ago
15

If 3.0×105 j of heat are added to the ice, what is the final temperature of the system?

Chemistry
1 answer:
lara [203]3 years ago
3 0
A wet-chemistry biochemical analyzer<span> was assessed for in-practice veterinary use. Its small size may mean a cost-effective method for low-throughput in-house biochemical analyses for first-opinion practice. The objectives of our study were to determine imprecision, total observed error, and acceptability of the </span>analyzer<span> for measurement of common canine and feline </span>serum<span> analytes, and to compare clinical </span>sample<span> results to those from a commercial reference </span>analyzer<span>. Imprecision was determined by within- and between-run repeatability for canine and feline pooled </span>samples<span>, and manufacturer-supplied quality control material (QCM). Total observed error (TEobs) was determined for pooled </span>samples<span> and QCM. Performance was assessed for canine and feline pooled </span>samples<span> by sigma metric determination. Agreement and errors between the in-practice and reference </span>analyzers<span> were determined for canine and feline clinical </span>samples<span> by Bland-Altman and Deming regression analyses. Within- and between-run precision was high for most analytes, and TEobs(%) was mostly lower than total allowable error. Performance based on sigma metrics was good (σ > 4) for many analytes and marginal (σ > 3) for most of the remainder. Correlation between the </span>analyzers<span> was very high for most canine analytes and high for most feline analytes. Between-</span>analyzer<span> bias was generally attributed to high constant error. The in-practice </span>analyzer<span> showed good overall performance, with only calcium and phosphate analyses identified as significantly problematic. Agreement for most analytes was insufficient for transposition of reference intervals, and we recommend that in-practice-specific reference intervals be established in the laboratory.</span>
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In class, it was mentioned that at 25 °C, the pH of pure water is 7.00 becuase Kw = 1.0 × 10−14. At other temperatures, the pH v
ivanzaharov [21]

Answer:

Kw = 2.88 × 10⁻¹⁵

Explanation:

Let's consider the dissociation of water.

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

The equilibrium constant Kw is:

Kw = [H⁺].[OH⁻]

If pH = 7.27, we can find [H⁺]:

pH = -log [H⁺]

H⁺ = anti log (-pH) = anti log (-7.27) = 5.37 × 10⁻⁸ M

According to the balanced equation, 1 mole of H⁺ is produced per mole of OH⁻. So, [H⁺] = [OH⁻] = 5.37 × 10⁻⁸ M

Then,

Kw = [H⁺].[OH⁻]= (5.37 × 10⁻⁸)² = 2.88 × 10⁻¹⁵

5 0
3 years ago
Temperature has no significance during the calculation of numerical .why?? <br>^_^​
Brums [2.3K]

Answer:

Explanation:

concepts, such as the internal energy of a system; heat or sensible heat, which are defined as types of energy transfer (as is work); or for the characteristic energy of a degree of freedom in a thermal system {\displaystyle kT}kT, where {\displaystyle T}T is temperature and {\displaystyle k}k is the Boltzmann constant.

7 0
3 years ago
Water (2350 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the ini
Eva8 [605]

Answer:

40.7062 °C  

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C ×ΔT

Where,  

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g  °C

m is the mass of water

ΔT = (100 - x) °C  

Given,

Mass = 2350 g

Q = 5.83 × 10⁵ J

Applying the values as:

Q = m C ×ΔT

5.83 × 10⁵ = 2350 × 4.184 × (100 - x)

<u>x, Initial temperature = 40.7062 °C  </u>

3 0
3 years ago
Compute the value of the molar heat capacity at constant volume, CVCV, for CO2CO2 on the assumption that there is no vibrational
katovenus [111]

Answer:

Explanation:

Molar heat capacity at constant volume Cv  of a gas = n x .5 R where n is degree of freedom of the gas molecules

CO₂ is a linear molecule , so number of degree of freedom =  3 + 2 = 5

3 is translational and 2 is rotational degree of freedom . There is no vibrational degree of freedom given .

So Cv = 5 / 2 R

= 2.5 R .

4 0
3 years ago
A 4o kg skier starts at the top of a 12 meter high slope at the bottom she is traveling 10 m/s how much energy does she lose to
allsm [11]
I think the answer is b because that is true 
5 0
3 years ago
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