Answers
Answer : The heat energy needed would be, 6486.5125 J
Explanation :
To calculate the change in temperature, we use the equation:
where,
q = heat needed = ?
m = mass of aluminum = 223 g
c = specific heat capacity of aluminum =
= change in temperature
= initial temperature =
= final temperature =
Putting values in above equation, we get:
Therefore, the heat energy needed would be, 6486.5125 J
Related Questions
1. At room temperature, air is usually a (solid, liquid, gas).
The resolution calculated for the tlc separation of a two-component mixture is determined to be equal to 2.0. does the represent good separation or not? explain
Which chemical equation follows the law of conservation of mass?
Draw an arrow-pushing mechanism for the elimination reaction between NaOH in ethanol and each of the following haloalkanes: (E1/E2)(i) 1-bromobutane;(ii) 2-bromo-2-methylpentane.
Answers
Answer:
Newton’s law of inertia is illustrated in tests with crash dummies, seat belts, and airbags, wherein the object stays in motion unless there is an unbalanced force applied to it.
Inertia is the main reason why there are seatbelts and airbags in the car. In this case, when the seatbelt is trapped to the passenger, the passenger experiences the same state of motion as the car. If the car accelerates/decelerates, the passenger experiences it too. When the car experiences collision, an unbalance force is acted upon it. This causes the car to stop abruptly, and the passenger shares the same state of motion because of the seatbelt and the airbags that apply the unbalanced force to stop the passenger to go forward.
3. Why do elements in the same group have similar chemical properties?
4.What are valence electrons?
5.Write the symbol for each of the following elements:
The halogen in period three
The alkali metal in period two
The noble gas in period one
The alkaline earth element in period six
Any transition metal in the 5th period
A metaloid in group 14
A nonmetal in group 16
6. What are two differences between a metal and a nonmeta
PLEASE ASAP!!!!!!!!!!!!!!
Answers
Answer:
question 6. answer is Metals are considered to be electropositive in nature due to their ability to donate electrons. Whereas non-metals are electronegative as they generally accept electrons
Answers
Answer:
(a) The rate constant is 3.61×10^-3 s^-1
(b) 7.12×10^4 s
Explanation:
(a) Log (K2/K1) = Ea/2.303R × [1/T1 - 1/T2]
K1 = 8.33×10^-6 s^-1
Ea = 245 kJ = 245,000 J
R = 8.314 J/mol.K
T1 = 427°C = 427+273 = 700 K
T2 = 545°C = 546+273 = 818 K
Log (K2/8.33×10^-6) = 245,000/2.303 × [1/700 - 1/818]
Log (K2/8.33×10^-6) = 2.637
K2/8.33×10^-6 = 10^2.637
K2 = 8.33×10^-6 × 433.51 = 3.61×10^-3 s^-1
(b) The relationship between temperature and the time required for reactants to be consumed is inverse
t2 = T1t1/T2
T1 = 427 °C = 700 K
t1 = 8.32×10^4 s
T2 = 545 °C = 818 K
t2 = 700×8.32×10^4/818 = 7.12×10^4 s
Answers
Answer:
Explanation:
When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.
this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.
Answers
Answer:
Cellulose is a major component of tough cell walls that surround plant cells, and it's what makes plant stems, leaves, and branches so strong. ... This forms long, cable-like structures, which combine with other cellulose molecules and is what produces such a strong support structure.
Explanation:
Answer:
Cellulose, a tough, fibrous, and water-insoluble polysaccharide, plays an integral role in keeping the structure of plant cell walls stable.
Explanation:
Cellulose chains are arranged in microfibrils or bundles of polysaccharide that are arranged in fibrils (bundles of microfibrils), which in turn make up the plant cell wall.
Answers
Answer:
Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.
Explanation:
A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).
Epoxidation can occur at both face of double bond result in formation of two stereoisomers.
Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane
Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.