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3 years ago

9

Make a prediction about the relationship between electrons and molecular shapes

1 answer:

7nadin3 [17]3 years ago

4 0

The electrons and the nuclei will settle into positions that minimize repulsion and maximize attraction.

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Pls help me on this question????

astraxan [27]

Explanation:

The first one is series circuit as it is joined one after another while the second one is parallel circuit joined in a parallel way. Ok the line just like this --------- is the wire , the sign like this | | is the battery and the circles with cross are the bulbs

Hope this helps :) Have a great day

7 0

3 years ago

What is the percent by mass of oxygen in carbon dioxide (CO2)?

wariber [46]

Answer:

= 72.73%

Explanation:

The percentage by mass of an element is given by;

% element = total mass of element in compounds/molar mass of compound × 100

The mass of oxygen in carbon dioxide = 32 g

Molar mass of CO2 = 44 g

Therefore;

% of O2 = 32/44 × 100%

<u>= 72.73%</u>

8 0

3 years ago

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

So

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

So

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

3 0

3 years ago

Question 6<br> How many molecules are in 4.48 liters of CO 2?

Dovator [93]

22.4 molecules are in 4.48 liters of CO 2

8 0

3 years ago

If a sample originally has 1,000 atoms and of carbon-14 how many would remain after 17000 years

Triss [41]

Answer:

12.791

Explanation:

6 0

2 years ago