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Kruka [31]
4 years ago
6

Methylhydrazine, CH6N2, is commonly used as a liquid rocket fuel. The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mo

l. How much heat is released when 206.2 g of methylhydrazine is burned? If appropriate, express your answer in scientific notion. (Click on the answer box to show the pallet.)
Chemistry
1 answer:
Gelneren [198K]4 years ago
7 0

Answer:

-5.82\times 10^3\ kJ

Explanation:

Given that:

The heat of combustion of methylhydrazine = -1.30\times 10^3\ kJ/mol

It means that:

1 mole of methylhydrazine on combustion, releases 1.30\times 10^3\ kJ of energy

Molar mass of ‎methylhydrazine = 46.072 g/mol

Means that, 1 mole of methylhydrazine contains 46.072 g of methylhydrazine

Thus,

46.072 g of methylhydrazine on combustion, releases 1.30\times 10^3\ kJ of energy

Also,

1 g of methylhydrazine on combustion, releases \frac {1.30\times 10^3}{46.072}\ kJ of energy

206.2 g of methylhydrazine on combustion, releases \frac {1.30\times 10^3}{46.072}\times 206.2\ kJ of energy

Heat released = -5.82\times 10^3\ kJ

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It is present in both solid and gas form.

It has a piney odor.

The physical appearance of toxaphene is yellow in colour amber waxy form.

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In 1940 it was manufactured.

It is a different composition of 670 chemicals.

According to questions,

Given,

Mass of toxaphene = 10^-5 gm

Mass of soil = 100 gm

Toxaphene soil concentrations = 10^-5 / 100 = 10^-7 gm/gm

1 gm/gm = 109 ppb

Concentration in units of ppb =10^-7 * 109 ppb = 100 ppb

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1 year ago
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MrRissso [65]

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8 0
3 years ago
40.0g of NaOH is dissolved in water. How much heat (in joules) was released?
adelina 88 [10]
<span>The molar heat of solution of NaOH is -445,100 J/mol. To compute much heat (in J) will be released if 40.00 g of NaOH are dissolved in  water, we first convert the given grams of NaOH to moles of NaOH, and use the given molar heat of solution to compute for the energy. (Using dimensional analysis):

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4 0
3 years ago
You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration
vovangra [49]

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM



7 0
4 years ago
Round to 3 significant figures.<br> 1.235
tigry1 [53]

Answer:

3 SigFigs = 1.24

1.24 x 10^{0}

Explanation:

3 0
4 years ago
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