Answer:
waxing is in the evening and waning is in the morning if you cant see the moon in the evening it is waning if you can see it in the evening it is waxing
D. 1,1,2,1
<h3>Further explanation </h3>
Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction(unbalanced)
Zn+K₂CrO₄ ⇒ K + ZnCrO₄
Give coefficient
aZn+K₂CrO₄ ⇒ bK + cZnCrO₄
Zn, left=a, right=c⇒a=c
K, left=2, right=b⇒b=2
Cr, left=1, right=c⇒c=1⇒a=1
O,left=4,right=4c⇒4c=4⇒c=1
Reaction(balanced0
Zn+K₂CrO₄ ⇒ 2K + ZnCrO₄
Answer:
Mass = 5.56 g
Explanation:
Given data:
Mass of Cl₂ = 4.45 g
Mass of NaCl produced = ?
Solution:
Chemical equation:
2Cl₂ + 4NaOH → 3NaCl + NaClO₂ + 2H₂O
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.45 g/ 71 g/mol
Number of moles = 0.063 mol
Now we will compare the moles of Cl₂ with NaCl.
Cl₂ : NaCl
2 : 3
0.063 : 3/2×0.063 =0.095 mol
Mass of NaCl:
Mass = number of moles × molar mass
Mass = 0.095 mol × 58.5 g/mol
Mass = 5.56 g
Answer:
A.......
Explanation: The sun is the main source of energy
Answer:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
