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2 years ago

What happens when magma cools during the rock cycle?

Chemistry

1 answer:

dalvyx [7]2 years ago

5 0

When magma cools down it turns into rocks these rocks are known as extrusive igneous rock

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Which image is cohesion stronger than adhesion and which image is adhesion stronger than cohesion

matrenka [14]

Image C is adhesion stronger and Image D is cohesion stronger

5 0

2 years ago

A sample of oxygen gas has a volume of 150.0 ml when its pressure is .947 atm. what will the volume of the gas be if the tempera

joja [24]

D
Hope it helped you get an A, Kayla.

7 0

2 years ago

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

6 0

2 years ago

Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10

PtichkaEL [24]

Answer:

$K.E.=1.97\times 10^{-21}\ J$

Explanation:

Given that:-

Pressure = $6.9\times 10^5\ Pa$

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

Given the value of pressure = 43,836 Pa

So,

$6.9\times 10^5\ Pa$ = $\frac{6.9\times 10^5}{101325}$ atm

Pressure = 6.80977 atm

Volume = $2.3\times 10^{-3}\ m^3$ = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

So, $K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J$

$K.E.=1.97\times 10^{-21}\ J$

3 0

2 years ago

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

$A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}$

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

$A=\text{ 50 }\times2^{\frac{-t}{14.3}}$

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

$\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}$

So, after 84 days the P-32 remaining will be 0.85 mg

4 0

10 months ago