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Alexxandr [17]
3 years ago
13

Ionic compounds are compounds made of?

Chemistry
1 answer:
Llana [10]3 years ago
3 0

Answer:

a

Explanation:

it made of metals and non metals

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Can someone help me
noname [10]
From Earth's<span> density we can estimate what elements must compose the </span>Earth; an iron core<span> just happens to estimate </span>Earth's<span> mass the best. Now from energy waves, geologists use seismometers to measure movements in </span>Earth's<span> interior (e.g. Earthquakes), These energy-waves form compressional and shear waves</span>
4 0
3 years ago
Hund's rule states that electrons must spread out within a given subshell before they can pair
Temka [501]

Answer:

Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.

Explanation:

If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.

6 0
3 years ago
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coat the metal preventing it from furt
Rasek [7]

Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

7 0
3 years ago
What is the part of an experiment that is not being tested and is used for comparison
STALIN [3.7K]
The controlled variable, I think.
7 0
3 years ago
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