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3 years ago

Which of the following is true of solutes dissolving in water?

Chemistry

1 answer:

Talja [164]3 years ago

3 0

B is the answer to your question.

C2H4 is not capable of hydrogen bonding because the H's are attached to the Carbon, and the charge is 0.

Although HCl cannot hydrogen bond, that aspect does not hinder it's ability to dissolve. Because HCl is polar and so is water, the positive side of H2O will be attracted to the negative side of HCl, thus "tearing" the molecule apart. (Like dissolves like - polar dissolves polar)

Based on the Solubility rule, KBr is soluble because it contains a group 1 metal.

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Answer: Earth, though the dwarf planet Pluto does have ice on it.

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Determine a massa de um átomo de carbono12 em gramas

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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi

Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM → 54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

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3 years ago

Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys

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Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

$Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}$ .

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both $[\mathrm{SO_2\, (g)}]$ and $[\mathrm{O_2\, (g)}]$ will increase if the pressure is increased through compression. However, because $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient $Q$ .

As a result, the increase in pressure will have no impact on the value of $Q\!$ . If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

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3 years ago

Balance the equation xcl2(aq)+agno3(aq)=x(no3)2(aq)+agcl(s)

lisabon 2012 [21]

Explanation:

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