Answer:
Ok So....
Explanation:
The condition in which a body covers equal distance in an equal interval of the time is known to be uniform velocity. if both its magnitude and direction do not change with time then it can be said that the body is at uniform velocity. Hope this helps!~ (Sorry if you get it wrong or I did-)
Answer:
Many forms of energy exist, but they all fall into two basic categories:
Potential energyKinetic energy
Answer:
Heat required to melt 26.0 g of ice at its melting point is 8.66 kJ.
Explanation:
Number of moles of water in 26 g of water: 26× moles
=1.44 moles
The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol.
we have relation as:
q = n × ΔH
where:
q = heat
n = moles
Δ
H = enthalpy
So calculating we get,
q= 1.44*6.02 kJ
q= 8.66 kJ
We require 8.66 kJ of energy to melt 26g of ice.
Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ * = 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g