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3 years ago

Plz help! The best answer will be marked as brainliest!

Chemistry

2 answers:

Ann [662]3 years ago

7 0

Answer:

About this Design So you did you here the news going around? The elements Oxygen and Magnesium are finally together! OMG right?!

Explanation:

from google, NOT A MEME LOL

harina [27]3 years ago

6 0

I heard Magnesium and Oxygen were going out.

So I was like, "OMg!"

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Please help me fast!!!!

Alinara [238K]

Answer:

The answer is THEORY! Hope it helps!

3 0

4 years ago

Read 2 more answers

One of relatively few reactions that takes place directly between two solids at room temperature is Ba(OH)2 · 8H2O + ammonium th

Hatshy [7]

Answer:

$\boxed{\text{3.1 g}}$

Explanation:

We will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 315.46 76.12

Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O

m/g: 6.5

1. Moles of Ba(OH)₂·8H₂O

$\text{Moles of Ba(OH)$_{2}\cdot $8H$_{2}$O}\\= \text{ 6.5 g Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}{\text{ 315.46 g Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}$

2. Moles of NH₄SCN

The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

$\text{Moles of NH$_{4}$SCN} =\text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{2 mol NH$_{4}$SCN}}{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0412 mol NH$_{4}$SCN}$

3. Mass of NH₄SCN

$\text{Mass of NH$_{4}$SCN} = \text{0.0412 mol NH$_{4}$SCN} \times \dfrac{\text{76.12 g NH$_{4}$SCN}}{\text{1 mol NH$_{4}$SCN}} =\\\textbf{3.1 g NH$_{4}$SCN}\\\\\text{You must use }\boxed{\textbf{3.1 g}}\text{ NH$_{4}$SCN}$

5 0

4 years ago

Consider the following reversible reaction.

dmitriy555 [2]

Answer:

Keq = [CO₂]/[O₂]

Explanation:

Step 1: Write the balanced equation for the reaction at equilibrium

C(s) + O₂(g) ⇄ CO₂(g)

Step 2: Write the expression for the equilibrium constant (Keq)

The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:

Keq = [CO₂]/[O₂]

6 0

3 years ago

Read 2 more answers

Predict the ground-state electron configuration of the following ions. Write your answers in abbreviated form, that is, beginnin

abruzzese [7]

These are three questions and three complete answers

Answer:

a) Cr²⁺: [Ar] 4s² 3d²

b) Cu²⁺: [Ar] 4s² 3d⁷

c) Co³⁺: [Ar] 4s² 3d⁴

Explanation:

a) Cr²⁺

Z = 24

Number of protons: 24

Number of elecrons of the neutral atom: 24

Charge of the ions: + 2

Number of electrons of the ion: 24 - charge = 24 - 2 = 22.

Electron configuration:

Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

Hence, for 22 electrons you get:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

[Ar] 4s² 3d²

b) Cu²⁺

Z = 29

Number of protons: 29

Number of elecrons of the neutral atom: 29

Charge of the ion: + 2

Number of electrons of the ion: 29 - charge = 29 - 2 = 27.

Electron configuration:

Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

Hence, for 27 electrons you get:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

[Ar] 4s² 3d⁷

c) Co³⁺

Z = 27

Number of protons: 27

Number of elecrons of the neutral atom: 27

Charge of the ion: + 3

Number of electrons of the ion: 27 - charge = 27 - 3 = 24.

Electron configuration:

Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

Hence, for 24 electrons you get:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴

Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

[Ar] 4s² 3d⁴

6 0

4 years ago

I need help with my chemistry homework. SHOW WORK

Veronika [31]

1) The gold will occupy a volume of 2.99 dm³.

$\text{Volume = 17 200 g} \times \frac{\text{1 cm}^{3}}{\text{5.75 g}} = \text{2990 cm}^{3} = \textbf{2.99 dm}^{3}\\$