Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Answer:
H, B, G, D, E, C, A, F
explanation: the ones that are the deepest are the oldest
Answer:
3.18 (w/w) %
Explanation:
In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:
NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl
<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>
<em> </em>Moles of thiosulfate in the titration are:
0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃
Thus, moles of NaClO in the initial solution are:
4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO
As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:
2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>
As mass of bleach is 0.496g, mass percent is:
0.0158g NaClO / 0.496g bleach ₓ 100 =
<h3>3.18 (w/w) % </h3>
Answer:
The answer is: 0,13 moles of CO2
Explanation:
We use the formula PV=nRT. The conditions STP are 1 atm of pressure and 273K of temperature:
PV=nRT n=PV/RT
n=1 atm x 2,8 L/ 0,082 l atm/K mol x 273K
n= 0,125078173 mol
Answer:
products and reaction
Explanation:
Products
1. FeSO4 and Cu
2. ZnSO4 and Fe
3. CaCl2 and H2
Reactions
1. Fe + CuSO4 → FeSO4 + Cu
2. FeSO4 + Zn → ZnSO4 + Fe
3. 2HCl + Ca → CaCl2 + H2
