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eimsori [14]
3 years ago
7

What variables(s) are constant for charler's law?

Chemistry
1 answer:
igomit [66]3 years ago
6 0
Pressure is the constant in Charles' law. If the pressure is constant, then this law upholds as being true.
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When 8.5 g of methane (ch4) is burned in a bomb calorimeter (heat capacity = 2.677 × 103 j/°c), the temperature rises from 24.00
Scrat [10]
The molar mass of methane is 16 g/mol. The heat absorbed by the calorimeter is the sensible heat, which is the heat gained or lost during a temperature change.

Sensible heat = CΔT = (<span>2.677 kJ/°C)(27.08 - 24</span>°C)
<em>Sensible heat = 8.24 kJ</em>
8 0
3 years ago
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If 3.0×105 j of heat are added to the ice, what is the final temperature of the system?
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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>
3 0
3 years ago
Give the values of n, l and ml for: (a) each orbital in the 2p subshell, (b) each orbital in the 5d subshell.?
SCORPION-xisa [38]

Hey there!:

* For 2p subshell :

n = 2, l =1, ml = -1, 0, +1

* for 5d subshell,

n = 5, l = 2, ml = -2, -1, 0, +1, +2


Hope that helps!

7 0
3 years ago
How many moles of sodium chloride can react with 18.3 liters of fluorine gas at 1.2 atmospheres and 299 Kelvin?
My name is Ann [436]

Answer:

1.79 mol.

Explanation:

  • For the balanced reaction:

<em>2NaCl + F₂ → 2NaF + Cl₂. </em>

It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.

  • Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = 18.3 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (299 K).

∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.

  • Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:

<em><u>Using cross multiplication:</u></em>

2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.

??? mol of NaCl is needed to react with → 0.895 mol of F₂.

∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.

4 0
3 years ago
2. Which test for iron(II) ions is conclusive ​
krok68 [10]

Answer:

please brainlist answer

Explanation:

The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.

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