GRADEC

The combustion of magnesium creates so much energy so quickly that it is hard to measure its enthalpy directly using a simple ca

riadik2000 [5.3K]

I believe your answer is: What is the enthalpy of formation of magnesium oxide?
Hope this helps! :)

6 0

3 years ago

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Consider the following balanced equation:3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)If 24.2 moles of Na3PO4(aq) reac

Tomtit [17]

Answer:

69.7% is percent yield

Explanation:

Based on the reaction:

3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

2 moles of Na3PO4 react producing 6 moles of NaNO3.

As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:

24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =

72.6 moles of NaNO3

As there are produced 50.6 moles of NaNO3, percent yield is:

50.6 moles NaNO3 / 72.6 moles NaNO3 =

<h3>69.7% is percent yield</h3>

7 0

3 years ago

How many grams of water can be formed from the reaction of 8.76 grams of H2 with 10.5 liters of O2 (at STP) according to the bal

Kisachek [45]

Answer:

16.9g of H₂O can be formed

Explanation:

Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

8.76g * (1mol / 2.016g) = 4.345 moles

<em>Moles O₂:</em>

PV = nRT

PV/RT = n

P = 1atm at STP

V = 10.5L

R = 0.082atmL/molK

T = 273.15K at STP

n = 1atm*10.5L / 0.082atmL/molK*273.15K

n = 0.469 moles of oxygen

For a complete reaction of 4.345 moles moles of hydrogen are required:

4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant

Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:

0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.

The mass is -Molar mas H₂O = 18.01g/mol-:

0.938 moles * (18.01g/mol) =

<h3>16.9g of H₂O can be formed</h3>

8 0

3 years ago

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.

dybincka [34]

<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>

pH:- 5.4
pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

$Antilog (-6) = {10}^{ - 6} , \\ Antilog (0.6) = 4$

put the value in equation

${10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]$

7 0

2 years ago

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A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago