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NemiM
4 days ago
11

Sea arches and sea stacks provide evidence of weathering and erosion. In three to five sentences, explain the roles weathering a

nd erosion play in creating these landforms.(4 points)
Chemistry
1 answer:
Tems11 [854]4 days ago
4 0

Answer:

?

Explanation:

You might be interested in
At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?
castortr0y [927]
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
8 0
13 days ago
Read 2 more answers
Consider the following system at equilibrium:
VMariaS [1037]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.

6 0
3 days ago
If the density of carbon tetrachloride is 1.59 g/ml, what is the volume in l, of 4.21 kg of carbon tetrachloride
Tems11 [854]

Density is defined as the mass-to-volume ratio. The formula for density can be expressed as:

density = \frac{mass}{volume}    -(1)

The density for carbon tetrachloride is provided as 1.59 g/ml   (given).

The mass of carbon tetrachloride is 4.21 kg   (as given).

Since, 1 kg = 1000 g

Thus, 4.21 kg = 4210 g

Utilizing the values in formula (1):

1.59 g/mL = \frac{4210 g}{volume}

volume = \frac{4210 g}{1.59 g/mL}

volume = 2647.799 mL

Since, 1 mL = 0.001 L

Hence, 2647.799 mL = 2.65 L

The resulting volume of carbon tetrachloride is 2.65 L.


6 0
11 days ago
Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the
Tems11 [854]

Answer:

THE MOLAR MASS OF XCL2 IS 400 g/mol

THE MOLAR MASS OF YCL2 IS 250 g/mol.

Explanation:

We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

Number of moles = mass / molar mass

Molar mass = mass / number of moles.

For XCl2,

mass = 100 g

number of moles = 0.25 mol

Thus, molar mass = mass / number of moles

Molar mass = 100 g / 0.25 mol

Molar mass = 400 g/mol.

For YCl2,

mass = 125 g

number of moles = 0.50 mol

Molar mass = 125 g / 0.50 mol

Molar mass = 250 g/mol.

Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.

3 0
5 days ago
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
lorasvet [960]

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

m_{tot}=m_{fuselage}+m_{passangers}

Para el aluminio:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

y

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

donde:

  • L es longitud
  • D es diámetro
  • e es grosor

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

Para el acero (mismo procedimiento):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

5 0
14 days ago
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