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gayaneshka [121]
3 years ago
11

when magnesium is burned in air the mass of the product is greater than the mass of the solid magnesium burned. A student claims

that this is a violation of the Low of conservation of mass do you agree or disagree with the claim. justify your answer
Chemistry
2 answers:
Nikitich [7]3 years ago
5 0
Veg egverbygbe rbbnfghjufddfg
il63 [147K]3 years ago
4 0
No, this is not a violation of the Law of conservation because when Magnesium is burnt it reacts with oxygen to form magnesium oxide. This end product is going to weigh more than the solid magnesium due to the addition of oxygen.
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Give the values of n, l and ml for: (a) each orbital in the 2p subshell, (b) each orbital in the 5d subshell.?
SCORPION-xisa [38]

Hey there!:

* For 2p subshell :

n = 2, l =1, ml = -1, 0, +1

* for 5d subshell,

n = 5, l = 2, ml = -2, -1, 0, +1, +2


Hope that helps!

7 0
3 years ago
which industrial city would have fewer air pollution incidents related to temperature inversions one on the great plains or one
egoroff_w [7]
I'm not positive, but I believe it would be one near the Rocky Mountains.
3 0
3 years ago
How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
OLEGan [10]

Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>

<em />

We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

3 0
3 years ago
CH3 + HCl &lt;=&gt; CH3Cl + H2O
dmitriy555 [2]

Answer:

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ =  X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

4 0
3 years ago
(image attached) AP CHEM ACID BASE Is my answer correct? I will mark as brainliest if you answer me. I don't need an explanation
pychu [463]

Answer:

Your answer is correct based on what I remember from AP Chemistry  

Explanation:

3 0
2 years ago
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