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ValentinkaMS [17]

3 years ago

13

Enter an abbreviated electron configuration for magnesium: Express your answer in complete form, in order of increasing energy.

For example, [He]2s22p2 would be entered as [He]2s^22p^2.

2 answers:

stiks02 [169]3 years ago

6 0

Answer:

1s2, 2s2,2p6,3s2

Explanation:

harkovskaia [24]3 years ago

6 0

Answer:

[Ne]3s²

Explanation:

Mg

1s2 2s2 2p6 3s2 or [Ne]3s²

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4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

3 years ago

A sample of hydrated magnesium sulfate (MgSO4)

yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

18x/120 + 18x = 6.572/12.845

Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

6 0

3 years ago

Write the name for CH2.

finlep [7]

Answer:

Methylene. :)

6 0

2 years ago

Read 2 more answers

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

<u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

<em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

<u>5. Calculate the number of moles of HgO</u>

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

<u>6. Convert to mass</u>

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Svet_ta [14]

Answer:

$V_1=23.3~mL$

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

$C_1*V_1=C_2*V_2$

Now we can identify the variables:

$C_1=~1.475_M$

$V_1=~?$

$C_2=~0.1374_M$

$V_2=~250.0~mL$

If we plug all the values into the equation:

$1.475_M*V_1=0.1374_M*250.0~mL$

And we solve for $V_1$ :

$V_1=\frac{0.1374_M*250.0~mL}{1.475_M}$

$V_1=23.3~mL$

I hope it helps!

8 0

3 years ago