Answer:
The percentage efficiency of the electrical element is approximately 82.186%
Explanation:
The given parameters are;
The thermal energy provided by the stove element, 
= 3.34 × 10³ J
The amount thermal energy gained by the kettle, 
= 5.95 × 10² J
The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;
Therefore, we get;
The percentage efficiency of the electrical element, η% ≈ 82.186%.
Answer:
The presence of 1-2% ethanol as catalyst, suppresses the oxidation of chloroform with oxygen to give a poisonous gas called phosgene. ... Here glycerol acts as negative catalyst. Criteria or characteristics of catalysts. i. The mass and chemical composition of catalyst should remain unchanged at the end of the reaction.
Explanation:
Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
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1)CH3-CH(OH)-СН2-СН2-СН2-СН2-СН3---(H2SO4)--›CH3-CH=CH-CH2-CH2-CH2-CH3+H2O
2)2-methyl-l-cyclohexanol---(h2so4)--›CH2=C(CH3)-CH2-CH2-CH2-CH2-CH3+H2O
