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dimaraw [331]
4 years ago
9

A reaction that is used to propel rockets is N2O4(l) + 2N2H4(l) --> 3N2(g) + 4H2O(g). This reaction has the advantage that ne

ither product is toxic, so no dangerous pollution is released. When the reaction consumes 10.0 g of liquid N2O4, it releases 124 kJ of heat. What is the value of ΔH (in kJ per mole N2O4) for the chemical equation as written?
Chemistry
1 answer:
jarptica [38.1K]4 years ago
4 0

Answer:

-1140 kJ/mol.

Explanation:

Let's consider the following balanced equation:

N₂O₄(l) + 2 N₂H₄(l) ⇒ 3 N₂(g) + 4 H₂O(g)

When 10.0 g of N₂O₄ react 124 kJ of heat are released. By convention, when heat is released we use the negative sign. Since the molar mass of N₂O₄ is 92.0 g/mol, the enthalpy of reaction is:

\Delta H=\frac{-124kJ}{10.0gN_{2}O_{4}} .\frac{92.0gN_{2}O_{4}}{1molN_{2}O_{4}} =-1140kJ/molN_{2}O_{4}

You might be interested in
What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m
velikii [3]

Answer: The [OH^-] of a solution is 10^{-12} M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of HCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.0025\times 1000}{250}=0.01

pH or pOH is the measure of acidity or alkalinity of a solution.

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,

1 mole of HCl gives 1 mole of H^+

Thus 0.01 moles of HCl gives =\frac{1}{1}\times 0.01=0.01 moles of H^+

Putting in the values:

[H^+][OH^-]=10^{-14}

[0.01][OH^-]=10^{-14}

[OH^-]=10^{-12}

Thus the [OH^-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is 10^{-12} M

8 0
4 years ago
at a particulat temperature, a smaple of pure water has a Kw of 7.7x10-14. what is the hydroxide concentration of this sample
ycow [4]

Answer: The hydroxide concentration of this sample is 3.85\times 10^{-7}

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

K_w=[H^+]\times [OH^-]

7.7\times 10^{-14}=[H^+]\times [OH^-]

As [H^+]=[OH^-]

2[OH^-]=7.7\times 10^{-14}

[OH^-]=3.85\times 10^{-7}

Thus hydroxide concentration of this sample is 3.85\times 10^{-7}

4 0
3 years ago
ANSWER QUICK PLEASE AND THANK YOU
madreJ [45]

One oxygen atom shares two electron with two hydrogen atoms in this way water molecules are formed. The bond between oxygen and hydrogen is covalent bond.

<h3>What is covalent bond ?</h3>

An electron exchange that results in the formation of electron pairs between atoms is known as a covalent bond. Bonding pairs or sharing pairs are the names given to these electron pairs.

Covalent bonding is the stable equilibrium of the attractive and repulsive forces between atoms when they share electrons.

According to the amount of shared electron pairs, there are three different forms of covalent bonds. single covalent bond is one type of covalent bond. covalent double bond and covalent triple bond.

Thus, option B is correct .

To learn more about covalent bond follow the link ?

brainly.com/question/10777799

#SPJ1

8 0
2 years ago
Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5
asambeis [7]
Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X}  (assume: 0,40-X\approx0,40)\\  \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\  \\ pH=-log([H_3O^{+}]=2,29

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!


3 0
3 years ago
Help ASAP please! I don’t understand this at all.
Zepler [3.9K]
Y the way , what is ASAP
4 0
3 years ago
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