The liquid that is been dispensed during titration as regards this question is Titrant.
- Titration can be regarded as common laboratory method that is been carried out during quantitative chemical analysis.
- This analysis helps to know the concentration of an identified analyte.
- Burette can be regarded as laboratory apparatus.
It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation of liquid, especially when performing titration.
- The specifications is been done base on their volume, or resolution.
- The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.
Therefore, burrete is used in volumetric analysis.
Learn more at:
brainly.com/question/2728613?referrer=searchResults
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
Answer:
2510
Explanation:
Since there are 10 milimetres in 1 centimetre, 1 cm = 10mm
There are 10 times as many milimetres as centimetres.
251cm X 10 = mm
251cm = 2510mm
Answer:
<h2> 162g/mol</h2>
Explanation:
The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:
<em>Nicotine has the formula </em>
<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>
<em>Assume that all the atoms in nicotine are present as products </em>
<h2>Solution</h2>
To find the empirical formula you need to find the moles of C, H, and N in each of the compound.
- 1.0 mol of CO₂ has 1.0 mol of C
- 0.70 mol of H₂O has 1.4 mol of H
- 0.20 mol of NO₂ has 0.20 mol of N
Thus, the ratio of moles is:
Divide all by the smallest number: 0.20
Hence, the empirical formula is C₅H₇N
Find the mass of 1 mole of units of the empirical formula:
Total mass = 60g + 7g + 14g = 81g
Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.
Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.
Answer:
all cell arise from pre existing cell
Explanation:
