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ioda
3 years ago
5

whg doe sthe transfer of electrons from magnesium to oxygen cause the ions produced to attract eachother

Chemistry
1 answer:
almond37 [142]3 years ago
8 0
Oxygen gains two electrons when it bonds to form a complete outer shell and magnesium loses two electrons when bonding to gain its full outer shell. 

As electrons are negative, the oxygen (which gains electrons) will become negative and the magnesium (which loses electrons) will become positive.

The negative and positive ions will then attract to one another due to the magnetic pull of the positive and negative.
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A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask?
horsena [70]
A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask<span>The given are: </span>
<span><span>1.      </span>Mass = ?</span><span><span /></span>
<span><span>2.      </span>Density = 1298 g/L</span>
3.      Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L

<span>Formula and derivation: </span><span><span>
1.      </span>density = mass / volume</span> <span><span>
2.      mass </span>= density / volume</span>

<span>Solution for the problem: </span><span><span>

1. mass = </span></span> <span> 1298 g/L / 0.125 L = 10384g
</span>


8 0
3 years ago
A 12 gram piece of metal is heated to 300 °C from 100 °C with 1120 Joules of energy. What is the specific heat of the metal?
Sergeeva-Olga [200]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

7 0
3 years ago
Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0
3 years ago
If 156 grams of chromium react with an excess of oxygen, as shown in the balanced chemical equation below, how many grams of chr
Sunny_sXe [5.5K]

Answer:

=759.95 grams.

Explanation:

The molar mass of chromium is 51.9961 g/mol

Therefore the number of moles of chromium in 156 grams is:

Number of moles =mass/RAM

=156g/51.9961g/mol

=3 moles.

From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.

Therefore 3 moles of chromium produce:

(3×2)/4 moles =1.5 moles of chromium oxide.

I mole of chromium oxide has a mass of 151.99 g

Thus 1.5 moles= 1.5mole ×151.99 g/mol

=759.95 grams.

3 0
3 years ago
Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t
iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0             0

C                     +S           +S

E                       S             S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

C                     +S'           +S'

E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0
3 years ago
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