corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.37g
Explanation:
0.21M means ; 0.21mol/dm³
1dm³=1L , so we can say 0.21mol/L
if 0.21mol of Zinc nitrate is contained in 1L of water
x will be contained in 135mL of water
x= 0.21*135*10³/1
=0.02835moles
number of moles= mass/ molar mass
mass= number of moles *molar mas
molar mass of Zn(NO₃)₂=189.36 g/mol
mass= 0.02835 *189.36
mass=5.37g
Cells are the basic unit of structure in all organisms and also the basic unit of reproduction.
What is the this (^ that up there) a description of?
Heat would be required : 1,670 J
<h3>Further explanation</h3>
Given
mass of H₂O=5 g
Required
Heat to melt
Solution
The heat to change the phase can be formulated :
Q = m.Lf (melting/freezing)
Lf=latent heat of fusion
The heat of fusion for water at 0 °C : 334 J/g
Input given values in formula :
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
