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2 years ago

What type of chemical reaction is demonstrated by this equation? A+BX-AX+B

Chemistry

2 answers:

Nezavi [6.7K]2 years ago

8 0

Answer:

single replacement

Explanation:Look it up and got it from quizlet it should be right.

BlackZzzverrR [31]2 years ago

3 0

Answer:

d an acid - base reaction.

Explanation:

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20 points

9966 [12]

Answer:

I`m pretty sure it`s electrical tape.

Explanation:

8 0

3 years ago

How many kilograms are in 125 pounds

zhuklara [117]

Answer:

125 pounds = 56.6990463 kilograms

Explanation:

3 0

3 years ago

The isomerization of methylisonitrile to acetonitrile is first order in CH3NC. CH3NC(g) → CH3CN(g) The half life of the reaction

sasho [114]

Answer: The rate constant for the given reaction is $4.33\times 10^{-6}s^{-1}$

Explanation:

For the given chemical equation:

$CH_3NC(g)\rightarrow CH_3CN(g)$

We are given that the above equation is undergoing first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

The rate constant is independent of the initial concentration for first order kinetics.

We are given:

$t_{1/2}$ = half life of the reaction = $1.60\times 10^5s$

Putting values in above equation, we get:

$k=\frac{0.693}{1.60\times 10^5s}=4.33\times 10^{-6}s^{-1}$

Hence, the rate constant for the given reaction is $4.33\times 10^{-6}s^{-1}$

5 0

3 years ago

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

Use the mole concept to calculate the number of atoms that are in a 1.75-mol sample of CHCl3.

Fiesta28 [93]

This is how I got to that answer. Since we don't know how many atoms there are in a mole, we use the number 6.02 x 10^-23. Now, just plug in what you have in the equation:

1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms.

6 0

3 years ago