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Lyrx [107]
3 years ago
14

If we place the like poles of two magnets facing each other what will happen?

Chemistry
1 answer:
Leviafan [203]3 years ago
8 0
The answer to that is D
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What is the overall effect of adding a solute to a Solution
user100 [1]

Answer:

By adding the solute in to solution boiling point is increased while freezing point is decreased.

Explanation:

When solute in added into the solvent the boiling point of solvent increases for example,

Water is boiled at 100 °C, when sodium chloride is added its boiling point increased. Ions of salt interact with solvent and prevent the water molecules to escape from the surface and form gas molecules. In order to make it boiled solution must be heated above 100  °C.

But there is different case with freezing point. Freezing point is the state in which substance converted into the solid. At given temperature when solute is added into the solvent it prevent the formation of solid. It required time to decrease the temperature first and as the temperature is decreases solid is formed.

8 0
3 years ago
The molar mass of H2O is 18.01 g/mol. The molar mass of O2is 32.00 g/mol. What mass of H2O,in grams, must react to produce 50.00
N76 [4]
Answer is: mass of water is 56.28 grams.
Chemical reaction: 2H₂O → 2H₂ + O₂.
m(O₂) = 50.00 g.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 50 g ÷ 32 g/mol.
n(O₂) = 1.5625 mol.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · 1.5625 mol.
n(H₂O) = 3.125 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 3.125 mol · 18.01 g/mol.
m(H₂O) = 56.28 g.
7 0
3 years ago
Read 2 more answers
Combustion analysis of toluene, a common organic solvent, gives 3.52 mg of co2 and 0.822 mg of h2o. if the compound contains onl
IRISSAK [1]
<span>C7H8 First, lookup the atomic weight of all involved elements Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Then calculate the molar masses of CO2 and H2O Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now calculate the number of moles of each product obtained Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK. moles CO2 = 3.52 / 44.0087 = 0.079984 moles H2O = 0.822 / 18.01488 = 0.045629 Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen. moles C = 0.079984 moles H = 0.045629 * 2 = 0.091258 So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting 1: 1.140953 The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like) 1 ( 1 : 1.140953) = 1 : 1.140953 2 ( 1 : 1.140953) = 2 : 2.281906 3 ( 1 : 1.140953) = 3 : 3.422859 4 ( 1 : 1.140953) = 4 : 4.563812 5 ( 1 : 1.140953) = 5 : 5.704765 6 ( 1 : 1.140953) = 6 : 6.845718 7 ( 1 : 1.140953) = 7 : 7.986671 8 ( 1 : 1.140953) = 8 : 9.127624 That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
7 0
3 years ago
Determine the electron geometry (eg), molecular geometry (mg), and polarity of SO2. Determine the electron geometry (eg), molecu
Temka [501]

Answer:

See explaination

Explanation:

The electrons geometry shows the special distribution of the electrons around of the central atom of the molecule.

The molecular geometry shows the special distribution of the atoms that form the molecule.

Please kindly check attachment for further solution.

6 0
3 years ago
How to do them 11-15 most difficult ones for me
BigorU [14]
Hope this can help you

6 0
3 years ago
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