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3 years ago

How many grams of mgo are produced during an enthalpy change of -231 kj ?

Chemistry

1 answer:

SCORPION-xisa [38]3 years ago

3 0

The reaction for the formation of MgO(s):

2 Mg (s) + O2(g) -à 2MgO(s) ΔH = -601.24 kJ/mol

The enthalpy information is taken from: http://webbook.nist.gov/cgi/inchi?ID=C1309484&Mask=2

From the equation and with an enthalpy change of -231 kJ:

-231 kJ * 2 mol Mg * (1/-601.24 kJ/mol) = 0.76841 mol Mg

Then, with the molar mass of MgO = 40.3,

0.76841 mol Mg *(2 mol MgO/2 mol Mg)* 40.3 g/mol MgO = 30.967 g MgO

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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago

In a physical change of matter,

mel-nik [20]

Answer: C. no new substances are formed

In the physical change of matter, there is no new substance that is formed. It is only the appearance of the matter that is being changed and not its chemical composition. Cutting, tearing and grinding are only some of the examples that exhibit physical change.

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3 years ago

Read 2 more answers

How do I calculate the moles consumed in a vinegar titration?

Shtirlitz [24]

Answer: Use the titration formula

Explanation:

8 0

3 years ago

After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?

FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: Mg₃N₂

I hope it helps!

5 0

3 years ago

Given 5.0 grams of lead (II) nitrate and 3.0

natulia [17]

Answer:

1.70 g of NaNO₃

Explanation:

The balance chemical equation for given double displacement reaction is,

Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: Calculate moles of each reactant:

Moles = Mass / M.Mass

For Pb(NO₃)₂:

Moles = 5.0 g / 331.21 g/mol

Moles = 0.0150 moles

For NaI:

Moles = 3.0 g / 149.89 g/mol

Moles = 0.020 moles

Step 2: Calculate Limiting reagent as;

According to equation,

1 mole of Pb(NO₃)₂ reacts with = 2 moles of NaI

So,

0.0150 moles of Pb(NO₃)₂ will react with = X moles of NaI

Solving for X,

X = 2 mol × 0.0150 mol / 1 mol

X = 0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: Find out moles of NaNO₃ formed:

According to equation,

2 mole of NaI produced = 2 moles of NaNO₃

So,

0.020 moles of NaI will produce = X moles of NaNO₃

Solving for X,

X = 2 mol × 0.020 mol / 2 mol

X = 0.020 moles of NaNO₃

Step 4: Calculate Mass of NaNO₃ produced;

As, Moles = Mass / M.Mass

Or,

Mass = Moles × M.Mass

Putting values,

Mass = 0.020 mol × 84.99 g/mol

Mass = 1.70 g of NaNO₃

7 0

3 years ago