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3 years ago

Explain why CaCl2 is likely to have properties similar to those of CaBr2

Chemistry

1 answer:

trasher [3.6K]3 years ago

5 0

Answer:

Because both compounds (CaCl2 and CaBr2) contain elements (bromine and chlorine) from the same group (group 7)

Explanation:

Elements are organized into groups on the periodic table based on the number of valence electrons contained in their outermost shell. These elements in the same group i.e. same number of valence electrons, will behave in a similar manner chemically.

CaCl2 and CaBr2 are two compounds that contains elements (Chlorine and Bromine) from the same group, which is group 7. Elements in group 7 are called HALOGENS and have 7 valence electrons in their outermost shell.

Bromine (Br) and Chorine (Cl) are responsible for the similarity in the properties of CaCl2 and CaBr2 because the Chlorine and Bromine contained in them will cause them to react similarly and behave similarly in the presence of other compounds.

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In nuclear fusion, small fragments, like atomic nuclei, neutrons, and other particles, form when energy is released. TRUE FALSE

LiRa [457]

Answer:

TRUE

Explanation:

It is true because along with energy, nuclear fusion releases other particles as well. This can include neutrons or atomic nuclei. An example of this is when hydrogen isotopes (D + T) fuse together to create helium it also releases a neutron. Thus D+T --> He + n + energy.

Therefore the answer is TRUE.

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4 years ago

Brainliest if CORRECT 11

Klio2033 [76]

Answer:

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Explanation:

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6 0

3 years ago

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Mercury melts at 234.3 k. Convert this this melting point to degrees celcius

DerKrebs [107]

Answer:

2069.85°C

Explanation:

0 K − 273.15 = -273.1 °C

Then,

234 K − 273.15 = 2069.85 °C

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3 0

4 years ago

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The specific heat of a metal is 3.76 cal/g°C. How much heat energy will it take to heat a 25.0 g cylinder of the metal from 21.5

Rudiy27

Answer:

The correct answer is "5217 Cal".

Explanation:

The given values are:

Specific heat,

c = 3.76 cal/g°C

Mass,

m = 25.0 g

Initial temperature,

T₁ = 21.5°C

Final temperature,

T₂ = 77.0°C

Now,

The heat energy will be:

⇒ $Q=mc \Delta t$

On substituting the given values, we get

⇒ $=25.0\times 3.76\times (77-21.5)$

⇒ $=25\times 3.76\times 55.5$

⇒ $=5217 \ Cal$

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3 years ago

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

joja [24]

Answer:

6.24 x 10-3 M

Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

$Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}$

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change $x$ , we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

$5.2x10^{-5}=\frac{x*x}{0.749-x}$

Thus, we obtain a quadratic equation whose solution is:

$x_1=-0.00627M\\x_2=0.00624M$

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

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3 years ago

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