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spayn [35]
3 years ago
8

What is the formula for calcium chloride

Physics
2 answers:
Naddik [55]3 years ago
6 0

The correct answer is - CaCl2

The calcium chloride is a salt, an inorganic compound. Its formula is CaCl2, with Ca being calcium, Cl being chloride, and the number 2 representing the number of chloride molecules.

The calcium chloride is a white colored crystalline solid when it is at room temperature, and it is highly soluble in water, acetone, and acetic acid. It has a molar mass of 110.98 g/mol, density of 2.15 g/cm³, and melting point at 772 °C.

Gnom [1K]3 years ago
4 0

the formula is CaC12

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If two balls have the same volume,
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Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

  • 1. Ball A will have the greater density
  • 2. Ball C and Ball D have the same density.
  • 3. Ball Q will have the greater density.
  • 4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

  • Density = Mass/Volume.

For Case 1:

  • Va = Vb and Ma = 2Mb
  • D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
  • Therefore, the density of ball A,
  • D(a) = 2D(b).
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For Case 2:

  • Vc = 3Vd,

  • Vd = (1/3)Vc

  • Md = (1/3)Mc

  • D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

  • D(c) = D(d).

  • Therefore, ball C and D have the same density

For Case 3:

  • Vp = 2Vq and Mp = Mq
  • D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
  • Therefore, the density of ball P is half the density of ball Q
  • Therefore, ball Q has the greater density.

For case 4:

  • Mx = (1/2)My
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Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

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2 years ago
Who was the most famous member of the Underground Railroad?
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Answer:

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Explanation:

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Answer:

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Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

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Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

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