Use the periodic to fill in the numbers in the electron configurations shown below

For the reaction shown here, 3.5 mola is mixed with 5.9 molb and 2.2 molc. what is the limiting reactant?3a+2b+c→2d

Misha Larkins [42]

<span>For equation A + 3B + 2C ---> 2D, 1 mole of A will produce 2 moles of D 3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D 2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>

7 0

3 years ago

A 2.0 liter balloon at a temperature of 25°C contains 0.1 mol of oxygen and 0.4 mol of nitrogen. What is the partial pressure of

Anastaziya [24]

4.89 or B for Plato users

7 0

3 years ago

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What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

5 0

3 years ago

If the formula for potassium chlorate is KClO3 and the formula for magnesium fluoride is MgF2, then what isthe formula for magne

mr_godi [17]

The formula for magnesium chlorate is Mg(ClO3)2.

5 0

3 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago