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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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Pick up plastic, reduce waste, reduce pollutants
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3 years ago
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Which aldehyde is an intermediate in the reduction of ethyl benzoate with lithium aluminum hydride?
garri49 [273]

Answer:

tetrahedral aldehyde

Explanation:

  1. The reaction begins with a hydride nucleophile reacting with the ester carbonyl carbon to form the tetrahedral intermediate.
  1. The carbonyl reforms to produce an aldehyde with the loss of the alkoxide ion.
  2. The resulting aldehyde undergoes a subsequent reaction with a hydride nucleophile to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups.
  3. Therefore, the tetrahedral intermediate is protonated to produce a primary alcohol.
6 0
3 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
Given the unbalanced equation: N2(g) + H2(g) → NH3(g)
mario62 [17]

Answer:

                   Option-4 (3:2) is the correct answer.

Explanation:

Following steps are taken to balance the given unbalanced chemical equation.

Step 1: Write the unbalanced chemical equation,

                         N₂ +  H₂ →  NH₃

Step 2: Balance Nitrogen Atoms;

There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.

                         N₂ +  H₂ →  2 NH₃

Step 3: Balance Hydrogen Atoms;

Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.

                         N₂ +  3 H₂ →  2 NH₃

Now, the equation is balanced.

Step 4: Finding out mole ratios:

From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.

3 0
3 years ago
Metals and their ore​
Maksim231197 [3]

Answer:

Zinc

Zinc blende

(ZnS)

-----------------

Lead

Galena

(PbS)

-----------------

Mercury

Cinnabar

(HgS)

----------------

Copper

Copper glance

(Cu2S)

-------------------

Explanation:

Had to learn this for a test

Also, I plead the 5th

8 0
3 years ago
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