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3 years ago

Magnesium oxide is a binary ionic compound. From the formula MgO, how do you know that Mg is the metal?

Chemistry

1 answer:

hammer [34]3 years ago

8 0

Answer:

See below.

Step-by-step explanation:

Method 1:

You know that oxygen is a non-metal.

If the compound is ionic, the oxygen must be combined with a metal.

The Mg must be a metal.

Method 2:

In the formulas of ionic compounds, the positive ion (usually a metal) is written first.

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2 years ago

What’s the best answer

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3 years ago

A 65.0-gram sample of some unknown metal at 100.0° C is added to 177.1 grams of water at 24.5° C. The temperature of the water r

jeka94

B.) 0.900 J/ (°C × g).

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3 years ago

Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon d

Korolek [52]

Answer:The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

Explanation:

Let the molecular formula of the oxide of metal be $X_2O_y$

$X_2O_y+yCO\rightarrrow yCO_2+2X$

Mass of metal product = 1.68 g

Moles of metal X = $\frac{1.68 g}{55.9 g/mol}=0.03005 mol$

1 mol of metal oxide produces 2 moles of metal X.

Then 0.03005 moles of metal X will be produced by:

$\frac{1}{2}\times 0.03005 mol=0.01502 mol$ of metal oxide

Mass of 0.01502 mol of metal oxide = 2.40 g (given)

$0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g$

y = 2.999 ≈ 3

The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

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3 years ago

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

This value corresponds to iron. Thus answer is B.

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3 years ago