Answer:
Anything in the environment that causes a change is called a stimulus.
Explanation:
- 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.
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Answer:
By weight they have the same mass, but the number of atoms is different
Explanation:
The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
