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Anton [14]
2 years ago
13

Hydrogen is special because it can act like two groups, _____ and _____

Chemistry
1 answer:
Vedmedyk [2.9K]2 years ago
4 0

Answer: gas and alkaline earth metal

Explanation:

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Explain how the cell membrane controls the movement of materials into and out of the cell ?
sukhopar [10]
It's called simple diffusion, the small molecules without charges such as oxygen and carbon dioxide can flow through a plasma membrane without assistance and without expending energy. Other substances such as proteins, glucose and charged particles called ions cannot pass through the selectively permeable membrane.
5 0
3 years ago
Which two formulas represent compounds
SVEN [57.7K]

CO2 ; H20- They are the only ones that, on both sides, combined with another element and bonding of atoms

4 0
3 years ago
Which machine is the least simple? a bike a screw an inclined plane a lever
Ivenika [448]

Answer: A bike

A bike is composed of various simple machines. It has a wheel and axle as one component, and it also has screws to hold the various parts together, along with levers and pulleys that are connected to the pedals. So all of these simple machine concepts work together to help transport the rider from point A to point B.

The other answer choices of screw, inclined plane, and lever, are fairly simple machines that don't have many things going on at once compared to a bike.

4 0
3 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
2 years ago
You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans
OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

4 0
3 years ago
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