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Rainbow [258]
2 years ago
10

A student weighs out 0. 0422 g of magnesium metal. The magnesium metal is reacted with excess hydrochloric acid to produce hydro

gen gas. A sample of hydrogen gas is collected over water in a eudiometer at 32. 0°c. The volume of collected gas is 43. 9 ml and the atmospheric pressure is 832 mmhg. Using the experimentally collected data, calculate r and the percent error.
Chemistry
1 answer:
lutik1710 [3]2 years ago
3 0
  • The student weighs out 0.0422 grams of the metal magnesium, thus we can figure that the more's, the magnesium he used, is the mass of the magnesium over the more mass, which is 0.024422.
  • That is approximately 0.001758.
  • Furthermore, it claims that too much hydrochloric acid causes the metal magnesium to react, producing hydrogen gas.
  • The volume of collected gas is 43.9 cc, the mastic pressure is 22 cc, and a sample of hydrogen gas is collected over water in a meter.
<h3>Is it true that calculations made utilizing experimental and gathered data result in a percent error? </h3>
  • Consequently, we are aware that magnesium and chloride react.
  • We create 1 as the reaction ratio is 1:2.
  • The hydrogen and 1 are more.
  • Magnesium chloride is more.
  • Therefore, based on this equation, we can infer that the amount of hydrogen that would be created in this scenario is greater than the amount of magnesium present here, or 0.001758 more.
  • Among hydrogen, there is.
  • \Once we convert the temperature from 32 Celsius to kelvin, we can tell you that the temperature is actually about 5.15 kelvin.
  • The gas has a volume of 43 in m, which is equal to 0.0439 liter and indicates that the pressure is approximately 832 millimeter.
  • Mercury, which is 2 times 13332 plus ca, or roughly 110922.24 par, is a mathematical constant.
  • So, in this instance, we are aware that p v = n r t.
  • The r in this case equals p v over n t, thus we want to determine the r.
  • So p is 110922.24. The temperature is 305.15 and the V is 0.04  over the n is 0.001758.
  • Let's proceed with the calculations right now.
  • In this instance, you will discover that the solution is 9.077 times 10; that is all there is to it.

To learn more about Magnesium chloride reactions visit:

brainly.com/question/27891157

#SPJ4

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arlik [135]

Answer:

Q = 306 kJ

Explanation:

Given that,

Mass, m = 60 kg

Specific heat, c = 1020 J/kg°C

The temperature changes from 20°C to 25°C.

Let Q be the change in thermal energy. The formula for the heat released is given by :

Q=mc\Delta T

Put all the values,

Q=60\times 1020\times (25-20)\\\\Q=306000\ J\\\\or\\\\Q=306\ kJ

So, 306 kJ is the change in thermal energy.

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Zepler [3.9K]
We have that energy=specific heat * change in temperature * mass. Thus, we have the final temperature (22) minus the initial temperature (55) to equal -33 as our change in temperature. Our specific heat is in J/g*C, so we're good with that because g stands for grams and the aluminium is measured in grams. As there are 10 grams of aluminum, we have
10*(-33)*0.902=-298 ish as our final temperature

An exothermic reaction would release energy and would therefore lose heat itself, while an endothermic reaction would absorb energy and gain heat. Therefore, losing heat would be an exothermic reaction

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Answer:

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Explanation:

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When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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3 years ago
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