Answer:
54g of water
Explanation:
Based on the reaction, 1 mole of methane produce 2 moles of water.
To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:
<em>Molar mass CH₄:</em>
1C = 12g/mol*1
4H = 1g/mol*4
12g/mol + 4g/mol = 16g/mol
<em>Moles methane: </em>
24g CH₄ * (1mol / 16g) = 1.5 moles methane
<em>Moles water:</em>
1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O
<em>Molar mass water:</em>
2H = 1g/mol*2
1O = 16g/mol*1
2g/mol + 16g/mol = 18g/mol
<em>Mass water:</em>
3.0moles H₂O * (18g / mol) =
<h3>54g of water</h3>
<span>1 mole glucose gives 2 moles of ethanol
moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles
so moles of ethanol produced = 2* 13.32 = 26.64 moles
weight of ethanol 26.64 * 46.07
=1227.30 gm or 1.23 Kg</span>
Answer:
Molarity = 0.21 M
Explanation:
Moles <em>solute </em>(mol) = Volume <em>solution</em> (L) x Molarity <em>solution </em>(M)
0.56 mol NaCl = 2.7 L x M
M = 0.2074074074
Answer:
H2O<en<phen
Explanation:
The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.
Explanation:
The given data is as follows.
Concentration of standard NaOH solution = 0.1922 M
Let the original acid solution concentration be x.
= 0.1 x M
= 10.00 mL (given)
The reaction equation is as follows.
Concentration × Volume of 
= Concentration × Volume of NaOH
x = 1.314 M
Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.
