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liubo4ka [24]
3 years ago
10

Are The Reactions Of Ordinary Molecular Hydrogen Slow Or Rapid ? Why ?

Chemistry
1 answer:
olchik [2.2K]3 years ago
4 0
In nature reactions of ordinary molecular hydrogen are slow since it's  a diatomic molecule whose atoms are held together by very strong covalent bonds.The reaction rate of hydrogen varies depending on temperature and the properties of the reactants, for instance under high temperatures above 500°C hydrogen reacts vigorously and with fluorine it reacts explosively even under low temperatures
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Which of these is a chemical property of matter?
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A. Ability to burn

Explanation:

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Type the correct answer in each box.
Arturiano [62]

Answer:

5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2

Explanation:

balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.

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3 years ago
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is :

C_2H_5NH_2_(_a_q_)     +     H^+_(_a_q_)   ---      C_2H_5NH_{3(aq)}^+

concentration of C_2H_5NH_{2(aq) = 10%

10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL

= 2.22 M

number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

= 20 mL \times  2.22 M= 44*10^{-3} mole

Concentration of C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq) is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0
3 years ago
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