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3 years ago

Explain in words or using your diagrams how beryllium atoms would react with fluorine atoms.

Chemistry

1 answer:

Marysya12 [62]3 years ago

8 0

Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

Be = 2, 2

F = 2, 7

Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.

When Be loses two electrons it becomes isoelectronic with He;

Be → Be²⁺ + 2e⁻

Also, when fluorine gains an electron, it becomes isoelectronic with Ne;

F + e⁻ → F⁻

This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.

Hence;

Be²⁺ + 2F⁻ → BeF₂

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What’s the reaction type of aqueous magnesium sulfate + aqueous sodium carbonate?

IrinaK [193]

Answer: Double displacement reaction

Explanation:

Double displacement reaction is defined as the reaction where exchange of ions takes place.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The chemical reaction between aqueous magnesium sulfate and aqueous sodium carbonate is represented as:

$MgSO_4(aq)+Na_2CO_3(aq)\rightarrow Na_2SO_4(aq)+MgCO_3(s)$

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3 years ago

Calcium chromate, cacro4, has a ksp of 7.1×10−4. what happens when calcium and chromate solutions are mixed to give 0.0200 m ca2

Serjik [45]

The solubility equilibrium of $CaCrO_{4}$ :

[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006

Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.

8 0

3 years ago

At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

$P=X_{A}P_{A}+X_{B}P_{B}$

Here, $X_{A}$ is mole fraction of A, $X_{B}$ is mole fraction of B, $P_{A}$ is partial pressure of A and $P_{B}$ is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

$X_{A}+X_{B}=1$

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

$X_{octane}=1-X_{hexane}=1-0.58=0.42$

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

$P=X_{hexane}P_{hexane}+X_{octane}P_{octane}$

Putting the values,

$P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg$

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

4 0

3 years ago

DDT:

Assoli18 [71]

Answer:

d. it does all of that and even harms humans to some extent

let me know if its right

8 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago