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3 years ago

What else is produced during the replacement reaction of magnesium and hydrochloric acid?

Chemistry

2 answers:

Harman [31]3 years ago

8 0

MgCI2......................

sweet [91]3 years ago

5 0

It's a single- replacement reaction so Mg+ 2HCl ➡️ H2+ MgCl2

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If a sample of pure calcium (Ca) contains 4.69 x 1022 atoms of iodine, how many moles of calcium are in the sample?. A) 0.0779 m

timama [110]

If a sample of pure calcium 9Ca) contains 4.69 x 1022 atoms of iodine, there are 0.0779 moles of calcium in the sample. The correct answer will be A.

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3 years ago

Find ∆G for a cell whose potential is +0.24 V with 4 moles of electrons exchanged.

Setler [38]

We need to use the following formula
Δ $G= -nF E_{cell}$

$E_{cell}= 0.24V$
n= 4 moles
F= constant= 96500C/mol

let's plug in the values.

ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ

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3 years ago

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storchak [24]

The name or type of covalent bond in which electrons are shared unequally is a polar covalent bond.

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3 years ago

The conversion factor that is present in almost all stoichiometry calculations is called the ________ __________ .

Kruka [31]

Mole=mass/molecular mass (Mr)
mole=volume/ 22.4
Molecular amount: Mole=amount atoms/ Avogadro's number
Concentration: mole=volume x M

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3 years ago

The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample o

adoni [48]

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

$v = \sqrt{\frac{3 \times R \times T}{M} }$

where,

R: ideal gas constant

T: absolute temperature

M: molar mass of the gas

We can use the info of argon to calculate the temperature for both samples.

$T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K$

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

$v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s$

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3 years ago