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ladessa [460]
2 years ago
9

10. What is the name of the acid with the formula HCIO?

Chemistry
1 answer:
pickupchik [31]2 years ago
3 0

Answer:

THE NAME OF THE ACID WITH FORMULA HCIO <em>IS </em>HYPOCHLOROUS ACID

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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

3 0
3 years ago
In your video respond to the following question/promt. Make sure you state a clear claim and explain with
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Answer:

Vvvbn n njjnjj

Explanation:

6 0
2 years ago
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Student Exploration: Limiting Reactants Vocabulary: chemical equation, chemical formula, chemical reaction, coefficient, limitin
SIZIF [17.4K]

Answer:

30 hot dogs

Explanation:

It is given that :

There are 4 packets of eight wieners, i.e. 4 x 8 = 32 wieners

There are 3 bags of ten buns, i.e. 3 x 10 = 30 buns

One hot dogs need 1 bun and 1 wiener to make a hot dog.

There are 30 buns, so 30 hot dogs can be made out by using all the 30 buns and the 30 wieners out of the 32 wieners.

Therefore, 30 hot dogs.

And the number of extra wieners left = 32 - 30 = 2 wieners.

6 0
2 years ago
Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

     = - ( -3 + log 1.948 )

     = 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0
1 year ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
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Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
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