1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³
Answer:
A model is developed for predicting oxygen uptake, muscle blood flow, and blood chemistry changes under exercise conditions. In this model, the working muscle mass system is analyzed. The conservation of matter principle is applied to the oxygen in a unit mass of working muscle under transient exercise conditions. This principle is used to relate the inflow of oxygen carried with the blood to the outflow carried with blood, the rate of change of oxygen stored in the muscle myoglobin, and the uptake by the muscle. Standard blood chemistry relations are incorporated to evaluate venous levels of oxygen, pH, and carbon dioxide.
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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