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3 years ago

How many molecules of carbon dioxide are in 9.080 x 10 ^-1 mol?

Chemistry

1 answer:

kakasveta [241]3 years ago

6 0

Answer:

5.46 8 x 10²³ molecules.

Explanation:

<em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:</u></em>

1.0 mole → 6.022 x 10²³ molecules.

9.08 x 10⁻¹ mole → ??? molecules.

∴ The no. of molecules of CO₂ are in 9.08 x 10⁻¹ mol = (6.022 x 10²³ molecules) ( 9.08 x 10⁻¹ mole) / (1.0 mol) = 5.46 8 x 10²³ molecules.

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Determine the freezing point depression of 2 kg of water when 2 mol salt is added to it. The kf of water is 1.86 degrees C/M. Sh

Svet_ta [14]

Answer: The freezing point depression is $1.86^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f$ = Depression in freezing point

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality = $\frac{\text {moles of solute}}{\text {mass of solvent in kg}}=\frac{2mol}{2kg}=1m$

$\Delta T_f=1.86mol/kg^0C\times 1m$

$\Delta T_f=1.86^0C$

Thus freezing point depression is $1.86^0C$

3 0

3 years ago

If you had an aqueous mixture that contained Ag+ , K+ , and Pb2+ cations, how many different solids could precipitate if a chlor

SCORPION-xisa [38]

Answer:

Two, KCl and PbCl₂.

Explanation:

Hello!

In this case, since the addition of chloride ions promote the following three ionic reactions:

$Ag^+(aq)+Cl^-(aq)\rightleftharpoons AgCl(s)\\\\K^+(aq)+Cl^-(aq)\rightleftharpoons KCl(aq)\\\\Pb^{2+}(aq)+Cl^-(aq)\rightleftharpoons PbCl_2(s)$

We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.

Best regards!

3 0

3 years ago

Consider the following Reaction.

jeyben [28]

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol

number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles

Molar mass of H3PO4 = 97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:

3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O

3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2

= (1/3)*0.5246

= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass

= 0.1749 * 262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%

Hope that helps!

3 0

3 years ago

How many pounds of ice are required to absorb 4900 kJ of heat as the ice melts? The heat of fusion of water is 0.334 kJ/g.

Goshia [24]

Answer:

m = 32.34 pounds of ice.

Explanation:

In this case we need to use the following expression of heat:

q = m * ΔHf (1)

Where:

q: heat absorbed in J or kJ

m: mass of the compound in g

ΔHf: heat of fusion of the water in kJ/g

We are asked to look for the mass of ice in pounds, so after we get the grams, we need to convert the grams to pounds, using the following conversion:

1 pound --------> 453.59 g (2)

So, we have the heat and heat of fusion, from (1) let's solve for the mass, and then, using (2) the conversion to pounds:

q = m * ΔHf

m = q / ΔHf

m = 4900 / 0.334 = 14,670.66 g of ice

Now, the conversion to pounds:

m = 14,670.66 g * 1 pound/453.59 g

<h2>m = 32.34 pounds of ice.</h2>

Hope this helps

5 0

3 years ago

A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen

Elenna [48]

Using the significant figure it would be 27.3

8 0

3 years ago