Answer:
Three of the five oxides are expected to form acidic solutions in water
Explanation:
We have different types of oxides : Acidic oxides, Basic oxides, Amphoteric oxides, Peroxides and Higher oxides.
Only acidic oxides will dissolve in water to give an acidic solution.
Considering the given oxides carefully,
- SO2 will dissolve in water to produce H2SO3 which is acidic.
- Y2O3 will dissolve in water to produce Yttrium(III) hydroxide which is basic.
- MgO will dissolve in water only to produce Mg(OH)2 which is also basic.
- Cl2O dichlorine mono oxide will dissolve in water to produce HClO which is acidic.
- N2O5 will dissolve in water to produce HNO3 which is also acidic.
To determine the pOH assuming water is the universal solvent take the value of 10 ^ -14 and then divide it by the hydronium concentration and then take the negative logarithm of the final answer that is the solution to the hydroxide ion concentration in the solution.
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).
See figure 1
I hope it helps!
Answer:
Explanation:
q = (mass) (temp change) (specific heat)
q = (10000 g) (40 °C) (0.385 J/g⋅°C) = 154000 J = 154 kJ
154 kJ / 2220 kJ/mol = 0.069369369 mol
0.069369369 mol times 44.0962 g/mol = 3.06 g (to three sig figs)
answer choice 4
<span>30.%
Determine the molar mass of NH3 and NO
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol
Moles NH3 = 8.5 / 17.03052 = 0.499103962 mol
Moles NO = 4.5 / 30.0057 = 0.149971505 mol
Looking at the balanced equation, for every mole of NH3 consumed, you should get one mole of NO. So if we had 100% yield, we should have 0.499103962 moles of NO. But we don't. The percent yield is a simple matter of division by what we did get by what we should get. So
0.149971505 / 0.499103962 = 0.300481497 = 30.0481497%
Rounding to 2 significant figures gives 30.% yield.</span>
