Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
C. (-2,-2) is the answer.
<u>Explanation:</u>
When we reflect a point (x, y) across the y-axis, after the reflection, the y-coordinate tends to be the same, however the x-coordinate is changed into its opposite sign.
Here U(2,-2) is reflected across the y-axis then,
the y-coordinate -2 remains the same and the x-coordinate is transformed into its opposite that is the sign of the x-coordinate will be changed as -2.
So the new coordinates of U after reflection will be (-2,-2).
There’s no pic for me to awnser your question
Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
