Please help me

The advantage of a fixed pulley on a flag pole is that it______.

Finger [1]

Your answer would be A for your homework

6 0

3 years ago

What are the possible orbitals for n = 3?

Tanzania [10]

Answer:The n = 3 shell, for example, contains three subshells: the 3s, 3p, and 3d orbitals. There is only one orbital in the n = 1 shell because there is only one way in which a sphere can be oriented in space.

Explanation:hope this helps:D

3 0

3 years ago

The drawing shows a chemical reaction between hydroxide and bromomethane to produce methanol and bromide ion. The middle drawing

Dafna11 [192]

Answer:

C.

Explanation:

Exothermic means the energy of products is lower than energy of reactants. Transition state will always be the highest energy among the two.

4 0

2 years ago

Describe the relationship between hydrogen oxygen and water

True [87]

Hydrogen is composed of H atom and oxygen is composed of O atom. For water, it is composed by both H and O atom. If you burn hydrogen in oxygen, you can get water. And if you electrolysis water, you can get hydrogen and oxygen.

6 0

4 years ago

Calculate the maximum solubility of silver carbonate, Ag2CO3 in g/L when in the presence of 0.057 M AgNO3. The solubility produc

Andreyy89

Answer:

Approximately $4.2 \times 10^{-7}\; \rm g \cdot L^{-1}$ .

Explanation:

Start by finding the concentration of $\rm Ag_2CO_3$ at equilibrium. The solubility equilibrium for

$\rm Ag_2CO_3 \; (s) \rightleftharpoons 2\, Ag^{+}\; (aq) + {CO_3}^{2-}\; (aq)$ .

The ratio between the coefficient of $\rm Ag_2CO_3$ and that of $\rm Ag^{+}$ is $1:2$ . For

Let the increase in $\rm {CO_3}^{2-}$ concentration be $+x\; \rm mol \cdot L^{-1}$ . The increase in $\rm Ag^{+}$ concentration would be $+2\,x\; \rm mol \cdot L^{-1}$ . Note, that because of the $0.057\; \rm mol \cdot L^{-1}$ of $\rm AgNO_3$ , the concentration of

The concentration of $\rm Ag^{+}$ would be $(0.057 + 2\, x) \; \rm mol\cdot L^{-1}$ .
The concentration of $\rm {CO_3}^{2-}$ would be $x\; \rm mol \cdot L^{-1}$ .

Apply the solubility product expression (again, note that in the equilibrium, the coefficient of $\rm Ag^{+}$ is two) to obtain:

$\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}$ .

Note, that the solubility product of $\rm Ag_2CO_3$ , $K_{\text{sp}} = 8.1 \times 10^{-12}$ is considerably small. Therefore, at equilibrium, the concentration of

Apply this approximation to simplify $(0.057 + x)^2\cdot x = 8.1 \times 10^{-12}$ :

$0.057^2\, x \approx (0.057 + x)^2 \cdot x = 8.1 \times 10^{-12}$ .

$\begin{aligned} x &\approx \frac{8.1 \times 10^{-12}}{0.057^2}\end{aligned}$ .

Calculate solubility (in grams per liter solution) from the concentration. The concentration of $\rm Ag_2CO_3$ is approximately $\displaystyle \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol\cdot L^{-1}$ , meaning that there are approximately $\displaystyle n = \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol$ of

$\begin{aligned}m &= n \cdot M \\ &\approx \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2} \; \rm mol\times 167.91\; g \cdot mol^{-1} \\ &\approx 4.2 \times 10^{-7}\; \rm g \end{aligned}$ .

As a result, the maximum solubility of $\rm Ag_2CO_3$ in this solution would be approximately $4.2 \times 10^{-7}\; \rm g \cdot L^{-1}$ .

8 0

3 years ago