Ask question

Ask question

All categories

2 years ago

8

An element belongs to group 4 and period 3. How many electrons does it have in its outermost shell and how many electronic shell

s does it have?

1 answer:

KATRIN_1 [288]2 years ago

3 0

Answer:

2 and 3

Explanation:

An element belongs to group 4 and period 3.

You might be interested in

What is the process of breaking down food so that it can be used by living things??

Darya [45]

I think the answer you're looking for is digestion.

4 0

3 years ago

A tank at is filled with of chlorine pentafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal

Ivan

Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

= 44.53 kPa

Total Pressure exerted by the gases = 60.58 kPa

Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

Mass = 13.3 g

Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

This is Dalton's law of Partial Pressure.

- Mole fraction of Chlorine Pentafluoride

= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

3 0

2 years ago

ANSWER FAST <br> is water wet? <br> a.) yes <br> b.) no<br> c.) a baby goat is a kid

Ksju [112]

Answer:

c

Explanation:

5 0

2 years ago

The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,

ludmilkaskok [199]

Answer: $4.3\times 10^{-13}s^{-1}$

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

Now put all the given values in this formula, we get

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

$(\frac{6.1\times 10^{-8}}{K_2})=141253.8$

Therefore, the value of the rate constant at 775.0 K is $4.3\times 10^{-13}s^{-1}$

5 0

2 years ago

When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2

Serjik [45]

Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
$\frac{P1}{T1} = \frac{P2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
$\frac{1100 bar}{T} = \frac{75.5 bar}{298 K}$
T = 4342 K
initial temperature was 4342 K

7 0

3 years ago