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qwelly [4]
1 year ago
5

A piece of magnesium ribbon reacts with oxygen to form magnesium oxide, mgo. what is the percent composition of the compound? qu

estion 1 options: 50% mg; 50% o 66% mg; 34% o 60% mg; 40% o 24% mg; 16% o
Chemistry
1 answer:
Wittaler [7]1 year ago
8 0

The Percent composition of  MgO is 60.304 % and 39.696 % respectively.

<h3>What percentage of the given chemical is there?</h3>

Mg + 1/2O₂ -------> MgO

We now want to know the percent composition of MgO in light of this reaction. Since the original reactants' and the product's mass are not provided, we can use the component's atomic weights to solve this problem.

Let's compute the molar mass of MgO since Mg has a molecular weight of 24.305 g/mol and O has a molecular weight of 15.999 g/mol:

molar mass of MgO= 24.305 + 15.999 = 40.304 \frac{g}{mol}

Now with this weight, let's see the percent composition of this compound:

Mg percentage  =\frac{24.305}{40.304 * 100}  = 60.304%

O percentage = \frac{15.999}{40.304 * 100}  = 39.696 %

The Percent composition of  MgO is 60.304 % and 39.696 % respectively.

Learn more about percent composition here:-

brainly.com/question/11885179

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Respuesta:

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Explicación:

Paso 1: Información provista

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Paso 2: Calcular el volumen final del gas

Si asumimos temperatura constante y comportamiento ideal, podemos calcular el volumen final del gas (V₂) usando la Ley de Boyle.

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4 0
3 years ago
PLEASE HELP! WILL GIVE BRAINLIEST:
Ad libitum [116K]

2.45 °C

From the Ideal gas law (Combined gas law)

PV/T = P'V'/T' .....eq 1

Where:

P - initial pressure

V - initial volume

T - initial temperature

P' - final pressure

V' - final volume

T' - final temperature.

To proceed we have to make T the subject from eq.1

Which is, T = P'V'T/PV.......eq.2

We have been provided with;

Standard temperature and pressure (STP)

P = 760 mm Hg (SP - Standard Pressure)

T = 273.15 K (ST - Standard Temperature)

V = 62.65 L

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V' = 78.31 L

T' = ? (what we require)

Therefore, we substitute the values into eq.2

T' =

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T' = 275.60 K

T = (275.60 - 273.15) ......To °C

T = 2.45 °C

>>>>> Answer

Have a nice studies.

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