What other objects are often arranged by their properties

How many electrons are in the highest occupied energy level of these atoms?

natima [27]

A. Barium with 56 is correct

7 0

3 years ago

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

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

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago

Element X has two isotopes. The table gives information about these isotopes.

Sedbober [7]

Answer : The average atomic mass of an element X is, 63.546 amu

Solution : Given,

Mass of isotope X-63 = 62.9296 amu

% abundance of isotope X-63 = 69.15% = 0.6915

Mass of isotope X-64 = 64.9278 amu

% abundance of isotope X-64 = 30.85% = 0.3085

Formula used for average atomic mass of an element X :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)]$

$\text{ Average atomic mass of an element X}=63.546amu$

Therefore, the average atomic mass of an element X is, 63.546 amu

5 0

4 years ago

Read 2 more answers

The volume of a gas is 1.15 L at STP. At what temperature would the volume of the gas increase to 1.56 L, if the pressure is dec

denis23 [38]

A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

<h3>What is an ideal gas?</h3>

An ideal gas is a gas whose behavior can be explained through ideal gas laws. One of them is the combined gas law.

A gas occupies 1.15 L (V₁) at STP (T₁ = 273,15 K and P₁ = 760 mmHg). We can calculate the temperature (T₂) at which V₂ = 1.56 L and P₂ = 650 mmHg, using the combined gas law.

$\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\\T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1} = \frac{650mmHg \times 1.56 L \times 273.15K}{760mmHg \times 1.15 L} = 317 K$

A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

Learn more about ideal gases here: brainly.com/question/15634266

#SPJ1

7 0

2 years ago

What is the latent heat of vaporization?

Ksivusya [100]

Answer:

option B

Explanation:

<h3>The energy required to go from liquid to gas / water vapour</h3>

6 0

3 years ago