The status of the state of matter of a material defines the phase. Pellet currently case 16 is defined; three of which solid, liquid and gas are referred to as conventional cases. There are a total of 16 state currently defined, these classic except; liquid crystal, amorphous solid, magnetic regularly, superconducting, superfluid, the Bose-Einstein condensation, Rydberg molecules, plasma (ionized gas), the quark-gluon plasma, degenerate matter, Superkate, the stringy liquid and SuperCam mine. Daily life in general, solid materials, include liquid or gaseous, but outside world, 99% of the material in the universe plasma (ionized gas) is in the form.
Number of moles of solute in 16.55 mL of 0.1475 M M2Cr3O7 is 0.00244 moles.
<u>Explanation:</u>
The properties and behavior of many solutions depend not only on the nature of the solute and solvent but also on the concentration of the solute in the solution. Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution:
,
Here ,
.
⇒ 
⇒ 
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Therefore, Number of moles of solute in 16.55 mL of 0.1475 M M2Cr3O7 is 0.00244 moles.
Answer:
1.195 M.
Explanation:
- We can calculate the concentration of the stock solution using the relation:
<em>M = (10Pd)/(molar mass).</em>
Where, M is the molarity of H₂SO₄.
P is the percent of H₂SO₄ (P = 40%).
d is the density of H₂SO₄ (d = 1.17 g/mL).
molar mass of H₂SO₄ = 98 g/mol.
∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.
- We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
<em>∴ (MV) before dilution = (MV) after dilution</em>
M before dilution = 4.78 M, V before dilution = 250 mL.
M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.
∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.
The sum of the masses of the reactants is equal to the sum of the masses of the products.
You must use 134 g O₂ to produce 118 g H₂O.


M_r: 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
Moles of H₂O = 150.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 8.374 mol H₂O
Moles of O₂ = 8.374 mol H₂O × (1 mol O₂/2 mol H₂O) = 4.187 mol O₂
Mass of O₂ = 4.1877 mol O₂ × (32.00 g O₂/1 mol O₂) = 134 g O₂