Samantha is using a lever to lift a rock. To life the rock most easily, where should she apply force to the lever

Pleasee help ASAP<br> will mark brainliest!

Anastasy [175]

Answer:

question 1- is B

question 2- is A

*I hope this helps you guys out!

7 0

2 years ago

Please help!

kiruha [24]

Answer:

The energy decreases as you move from the bottom to the top of the energy pyramid. 90% of the energy entering each step of the pyramid is used up by the consumers, only 10% of the energy get stored. So, when the next level on the pyramid eats them, that 10% gets passed down to them Use the diagram below to answer the question.

Explanation:

I am sorry if I am wrong

And can you give me brainliest please

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3 0

2 years ago

For the reaction below, Kp = 1.16 at 800.°C. CaCO3(s) equilibrium reaction arrow CaO(s) + CO2(g) If a 25.0-g sample of CaCO3 is

goblinko [34]

Answer:

76.0%

Explanation:

Let's consider the following reaction.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

At equilibrium, the equilibrium constant Kp is:

Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm

We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.16atm\times 14.4 L}{(0.08206atm.L/mol.K)\times 1073K} =0.190mol$

From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:

$0.190molCO_{2}.\frac{1molCaCO_{3}}{1molCO_{2}} .\frac{100.09gCaCO_{3}}{1molCaCO_{3}} =19.0gCaCO_{3}$

The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:

$\frac{19.0g}{25.0g} \times 100\%=76.0\%$

5 0

2 years ago

True or False - Changing the temperature will not affect the pressure of the mixture?

irina [24]

The answer would most likely be true.

7 0

3 years ago

Read 2 more answers

3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.

tiny-mole [99]

Answer:

The Avogadro's number is $N_A = 6.02289 *10^{23}$

Explanation:

From the question we are told that

The edge length is $L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}$

The density of the metal is $\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3$

The molar mass of Ba is $Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol$

Generally the volume of a unit cell is

$V = L^3$

substituting value

$V = [5.02 *10^{-10}]^3$

$V = 1.265*10^{-28}\ m^3$

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),

The volume of barium atom is

$V_a = \frac{V}{2} * 0.68$