Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
Choose any of these and it should be correct:
-atmospheric science
-environmental science
-climate science
-biology
-astronomy
-human physiology and medicine
Explanation:
I just had this question on edgenuity, this is what came up as correct. I hope this helps
Answer:
Explanation:
A solution of a weak base and its conjugate acid is a buffer.
The equation for the equilibrium is
The Henderson-Hasselbalch equation for a basic buffer is
![\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7Bp%7DK_%7B%5Ctext%7Bb%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BBH%7D%5E%7B%2B%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D)
Data:
[B] = 0.400 mol·L⁻¹
[BH⁺] = 0.250 mol·L⁻¹
Kb = 4.4 × 10⁻⁴
Calculations:
(a) Calculate pKb
pKb = -log(4.4× 10⁻⁴) = 3.36
(b) Calculate the pH
Answer:
a) But-1-ene
b) E-But-2-ene
c) Z-But-2-ene
d) 2-Methylpropene
Explanation:
In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u>
in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>
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For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>
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Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>
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See figure 1 to further explanations.
I hope it helps!
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