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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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Answer:

<h2>8.92 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 4.75 g

volume = 0.5325 cm³

We have

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We have the final answer as

<h3>8.92 g/cm³</h3>

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