Energy required to raise the temperature from 35°C - 45 °C= 25116 J.
specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.
solving,
Sample of liquid = 400. 0 g
temperature = 30. 0 ºc
joules of energy are required to raise the temperature of the water to 45. 0 ºc
therefore rise in temperature 45 - 30 = 15°C
Specific heat capacity = 4.186 J/g m °C
In kelvin = 273 + 15 = 288
= ∴ energy required = Q = m s ( t final - t initial)
= 400*4.186 * 15
= 25116 joule
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Explanation:
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We cannot see it its so small
Also yes it matches your hypothesis
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
It should be 1. 1.2 X 10^24
Explanation:
