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vladimir1956 [14]

3 years ago

6

CLIOT

1 answer:

zhuklara [117]3 years ago

6 0

The many electron diamonds around the central carbon are 1689

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5. How many moles are there in 222.3 grams of Ca(OH)2? (1) 1.5 mol (3) 5.5 mol (2) 3.0 mol (4) 22.3 mol 5

tensa zangetsu [6.8K]

Answer: (2)3.0 mol

Explanation: 222.3g= 222.3/74= 3.0 mol

1mol Ca(OH)2 is 74g

6 0

3 years ago

Between carbon and silicon, which element has the strongest attraction between its nucleus and its valence electrons? Explain?

Ede4ka [16]

Answer:

Between carbon and silicon, silicon element has the strongest attraction between its nucleus and its valence electrons.

because silicon's nucleus is bigger than carbon.silicon loses electrons easily than carbon.

8 0

3 years ago

A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and t

drek231 [11]

Answer:

(a) ΔU = 7.2x10²

(b) W = -5.1x10²

(c) q = 5.2x10²

Explanation:

From the definition of power (p), we have:

$p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t}$ (1)

<em>where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s) </em>

(a) We can calculate the energy (ΔU) using equation (1):

$\Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J$

(b) The work is related to pressure and volume by:

$\Delta W = -p \Delta V$

<em>where p: pressure and ΔV: change in volume = V final - V initial </em>

$\Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J$

(c) By the definition of Energy, we can calculate q:

$\Delta U = \Delta W + \Delta q$

<em>where Δq: is the heat transfer </em>

$\Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J$

I hope it helps you!

6 0

4 years ago

Help me please!!!!!!!!?????

kirill [66]

Answer:

i believe its A or D

Explanation:

sorry is its wrong

hope it helps

4 0

3 years ago

Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion requi

Firdavs [7]

Answer:

The concentration of lead ion required to just initiate precipitation is - $2.37\times10^-^5 M$

Explanation:

Lets calculate -:

Solubility equilibrium -: $PbI_2(s)$ ⇄ $Pb^2^+ (aq) + 2I^- (aq)$

Solubility product of $PbI_2$ , $Q=[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $=9.8\times10^-^9$

Concentration of $I^-$ $=[KI]=0.0492M$

When the ionic product exceeds the solubility product , precipitation of salt takes place .

$Q_s_p\geq K_s_p$

$[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $\geq 9.8\times10^-^9$

$[Pb^2^+]_i_n_i_t_i_a_l$ $[0.0492]^2$ $\geq 9.8\times10^-^9$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq \frac{9.8\times10^-^9}{[0.0492]^2}$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $\frac{9.8\times10^-^9}{2.42\times10^-^3}$

$[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $2.37\times10^-^5 M$

Thus , $PbI_2$ will start precipitating when $[Pb^2^+]_i_n_i_t_i_a_l$ $\geq 2.37\times10^-^5 M$ .

3 0

3 years ago