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3 years ago

A stable atom that has a large nucleus most likely contains

Chemistry

1 answer:

maria [59]3 years ago

8 0

Answer:

3. Equal numbers of protons and neutrons

Explanation:

The nucleus becomes unstable if the ratio of protons to neutrons is less than 1:1 or more than 1:1.5.

The most stable nucleus has a neutron proton ratio of 1:1 which means that they can not release a neutron or a proton to decay.

Nucleus 3 is therefore the most stable.

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If the pressure of a 2.00 L sample of gas is 50.0 kPa, what pressure does the gas exert if its volume is decreased to 20.0 mL?

dimulka [17.4K]

Answer:

P₂ = 5000 KPa

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 50.0 KPa

Final volume = 20.0 mL (20/1000=0.02 L)

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

50.0 KPa × 2.00L = P₂ × 0.02 L

P₂ = 100 KPa. L/0.02 L

P₂ = 5000 KPa

8 0

3 years ago

3. How does this difference in language relate to our propensity to save?

stich3 [128]

Answer:

If you speak a futureless language, you will speak about time identically meaning you will feel as they are identical, it increases your propensity to save.

Explanation:

hope this helps you :)

8 0

4 years ago

A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23

Anna35 [415]

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$