Ask question

Ask question

All categories

3 years ago

13

atomic radii decrease,moving from left to right across a period. As a result,the electrons become closer to the nucleus.What eff

ect does this movement have on the ionization energy (the amount of energy required to remove an electron from an atom)?

2 answers:

muminat3 years ago

4 0

Ionization energy increases.

IceJOKER [234]3 years ago

3 0

The ionization energy increases

You might be interested in

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can <u>put the values into the equation</u>:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can <u>solve for the initial temperature of gold</u>, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

3. How many molecules are in 0.500 moles of sulfur?

satela [25.4K]

Explanation:

No of molecules=0.500×6.023×10²³=3.011×10²³ molecules

7 0

3 years ago

A 48.0g sample of quartz, which has a specific heat capacity of 0.730·J·g−1°C−1, is dropped into an insulated container containi

Butoxors [25]

Answer:

The equilibrium temperature of the water is 26.7 °C

Explanation:

<u>Step 1:</u> Data given

Mass of the sample quartz = 48.0 grams

Specific heat capacity of the sample = 0.730 J/g°C

Initial temperature of the sample = 88.6°C

Mass of the water = 300.0 grams

Initial temperature = 25.0°C

Specific heat capacity of water = 4.184 J/g°C

<u>Step 2:</u> Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

Q = m(quartz)*c(quartz)*ΔT(quartz) = -m(water) * c(water) * ΔT(water)

⇒ mass of the quartz = 48.0 grams

⇒ c(quartz) = the specific heat capacity of quartz = 0.730 J/g°C

⇒ ΔT(quartz) = The change of temperature of the sample = T2 -88.6 °C

⇒ mass of water = 300.0 grams

⇒c(water) = the specific heat capacity of water = 4.184 J/g°C

⇒ ΔT= (water) = the change in temperature of water = T2 - 25.0°C

48.0 * 0.730 * (T2-88.6) -300.0 * 4.184 *(T2 - 25.0)

35.04(T2-88.6) = -1255.2 (T2-25)

35.04T2 -3104.544 = -1255.2T2 + 31380

1290.24T2 = 34484.544

T2 = 26.7 °C

The equilibrium temperature of the water is 26.7 °C

8 0

3 years ago

A plate moves 200 m in 10,000 years. What is its rate in cm/year?

Varvara68 [4.7K]

Answer:

The answer is

<h2>2 cm/year</h2>

Explanation:

To find the rate in cm/year we must first convert 200 m into cm

1 m = 100 cm

if 1 m = 100 cm

Then 200 m = 200 × 100 = 20 ,000 cm

So the rate is

<h2> $\frac{20000}{10000}$ </h2>

<u>Reduce the fraction with 10,000</u>

We have the final answer as

<h3>2 cm/year</h3>

Hope this helps you

3 0

3 years ago

Warm air holds _____.<br>

Agata [3.3K]

Answer:

less water than cl air

Explanation:

i think

8 0

3 years ago