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4 years ago

Which of the following would most likely act as a Bronsted-Lowry acid?

Chemistry

2 answers:

dlinn [17]4 years ago

7 0

Answer:

HCN

Explanation:

answered right on edge

levacccp [35]4 years ago

3 0

A Bronsted-Lowry acid is a proton donor (usually hydrogen ion). And a Bronsted-Lowry base is a proton acceptor (usually hydrogen ion). Consider a chemical reaction between HCl and NaOH. We have the reaction HCl + NaOH à NaCl + H2O. The hydroxide ions in the NaOH are bases because they accept hydrogen ions from acids to form water. And an acid produces hydrogen ions in solution by giving a proton to the water molecule. Therefore, the answer is d. a Bronsted-Lowry base.

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Calculate the number of moles of Cu in 3.8×10213.8×1021 atoms of Cu.

rosijanka [135]

Answer:

6.3×10⁻³ moles of Cu

Explanation:

The relation is, that 1 mol of particles are contained by 6.02×10²³ particles.

This is Avogadro's number that explains that: one mole determines the number of fundamental units that are contained in a constant number but they do not depend on the type of material or the type of particle, and this quantity is 6.02 × 10²³

We can make a rule of three:

6.02×10²³ atoms are contained 1 in mol of Cu

Therefore, 3.8×10²¹ atoms will be contained in (3.8×10²¹ . 1) / NA = 6.3×10⁻³ moles

7 0

4 years ago

Mass = 100g, volume=20ml, what is density?

Nonamiya [84]

Answer:

<h3>The answer is 5.0 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 100 g

volume = 20 mL

So we have

$density = \frac{100}{20} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

What is the pOH of a<br> 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

3 0

3 years ago

What main type of forces must be overcome between br2 when liquid br2 dissolves into ethanol?

Snezhnost [94]

Dipole-induced dipole forces.

In contrast to C2H5OH, which is a very polar molecule, bromine is non-polar and has zero dipole moment. Charges are separated within polar molecules.

When a nonpolar Br2 molecule interacts with a polar C2H5OH molecule, one half of the bromine molecule acquires a charge and the other half acquires an opposing charge. A dipole is created when two adjacent bromine molecules have different charges. Bromine molecules must thus overcome the intermolecular forces—also known as dipole-induced dipole forces—that form in order to dissolve in ethanol.

Find more on bromine at : brainly.com/question/865727

#SPJ4

8 0

2 years ago

For the reaction CO(g) + 3 H2(g) ⇌ H2O(g) + CH4(g) , Kc = 190 at 1000 K. If a vessel is filled with these gases such that the in

zheka24 [161]

Answer:

<u>the reaction will shift to the left </u>and produce more reactants

Explanation:

Step 1: data given

Kc = 190

Temperature = 1000 K

[CO] = 0.025 M

[H2] = 0.045 M

[H2O] = 0.025 M

[CH4] = 0.046 M

When Q=Kc, the system is at equilibrium and there is no shift to either the left or the right.

When Q<Kc, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

When Q>Kc, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants

Step 2: The balanced equation

CO(g) + 3 H2(g) ⇌ H2O(g) + CH4(g)

Step 3: calculate Q

Q = [H2O][CH4] / [CO][H2]³

Q = (0.025*0.046) / (0.025*0.045³)

Q = 504.8

Q > Kc

This means there are more products than reactants. To decrease the amount of products, <u>the reaction will shift to the left </u>and produce more reactants

8 0

3 years ago