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Pani-rosa [81]
3 years ago
7

Sodium metal reacts with water to produce hydrogen gas according to the following equation: 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Th

e product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 754 mm Hg. If the wet H2 gas formed occupies a volume of 8.77 L, the number of moles of Na reacted was mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.
Chemistry
1 answer:
Maru [420]3 years ago
3 0

Answer:

Number of moles of sodium reacted = 0.707 moles

Explanation:

P(H₂) = P(T) – P(H₂O)

P(H₂) = 754 – 17.5 = 736.5 mm Hg

Use the ideal gas equation which

PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature

<u>Re- arrange to calculate the number of moles and using the data provided</u>

n = P x V/R x T

n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)

n = 0.35348668

n = 0.353 moles H₂

<u>from the equation we know that</u>

0.353 mole H₂ x 2mole Na/1mole H₂, So

0.353 x 2 = 0.707 mole Na

The number of moles of Sodium metal reacted were 0.707 moles.

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Nikolay [14]
First; use the relationship of molarity and moles/liter to find the moles of the solution.
so 0.600M ×0.030L = 0.018 moles
Then use the mole to mole ratio of lithium to lithium carbonate
0.18 × (2 Li ÷1 Li2CO3) = 0.036
and then multiply by Avogadro's number to find the ions of lithium
0.036 moles × (6.022×10^{23} ) = 2.167 ×10∧22 ions Li
4 0
3 years ago
What is the simple sugar resulting from the hydrolysis of lactose?
garri49 [273]

Glucose and galactose (both simple sugars) are produced from the hydrolysis of lactose.

5 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ
egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

<u>Step 1:</u> Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

<u>Step 2:</u> Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+  in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed  in the back titration with Ca2+:

<u>Step 3: </u>Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

<u>Step 4:</u> Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

<u>Step 5:</u> Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction  with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

<u>Step 6:</u> Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

<u>Step 7: </u>Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

4 0
3 years ago
In which block of the periodic table should iron be placed
Dmitry_Shevchenko [17]
Iron should be in the 3d block

8 0
3 years ago
From the mass of Al reacted, how many moles are present? What is the mole ratio between Al and H2 in the balanced chemical react
tigry1 [53]

Answer:

From the mass of Al reacted, how many moles are present? We don't have the mass of Al reacted

What is the mole ratio between Al and H2 in the balanced chemical reaction?

By stoichiometry, ratio between Al and H₂ is 3:2

What units are required on the temperature when doing calculations with gases? °K degrees

The value of R is 0.082 L.atm/mol.K

Explanation:

A chemical reaction for this can be this one:

2Al (s)  + 6HCl (aq)  →  2AlCl₃(aq)  + 3H₂ (g)

Balanced

To get the moles which are present in the mass of Al reacted you must calculate:

Al Mass / Al molar mass = Al moles

The Ideal Gas equation is this one:

Pressure . Volume = n° moles . R (gases constant) . T°K

The temperature must be in K, because the units of R

We usually use R = 0.082 L.atm/mol.K

5 0
3 years ago
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