Given:
128g sample of titanium
2808J of heat energy
specific heat of titanium is 0.523 J/ g °C.
Required:
Change in temperature
Solution:
This can be solved
through the equation H = mCpT
where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.
Plugging in the
values into the equation
H = mCpT
2808J = (128g) (0.523
J /g °C) T
T
= 41.9 °C
Answer:
3.69 g
Explanation:
Given that:
The mass m = 325 g
The change in temperature ΔT = ( 1540 - 165)° C
= 1375 ° C
Heat capacity = 0.490 J/g°C
The amount of heat required:
q = mcΔT
q = 325 × 0.490 × 1375
q = 218968.75 J
q = 218.97 kJ
The equation for the reaction is expressed as:
Then,
1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.
Thus, for 218.97 kJ, the amount of ethyne gas required will be:
= 3.69 g
When the Earth's surface absorbs the sun's energy, it turns the light into heat. This heat on Earth's surface warms the air above it. The air over the Equator gets warmer than the surface air near the poles. This moving air is what we call wind.
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When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now write the chemical equation
CH₃COOH + CH₃OH ⇄ CH₃COOCH₃ + H₂O
Here
CH₃COOH is Ethanoic acid (Acetic acid)
CH₃OH is Methanol (Methyl Alcohol)
CH₃COOCH₃ is Methyl ethanoate (Methyl Acetate)
H₂O is water
Thus from the above conclusion we can say that When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.
Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427
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Answer: 0.52 L of 15 M will be used to prepare this amount of 0.52 M base.
Explanation:
But on diluting the number of moles remain same and thus we can use molarity equation.
(to be prepared)
where,
=concentration of stock solution = 15 M
= volume of stock solution = ?
= concentration of solution to be prepared = 0.52 M
= volume of solution to be prepared = 15 L
Thus 0.52 L of 15 M will be used to prepare this amount of 0.52 M base