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Alisiya [41]
3 years ago
15

What would be a correct unit of measurement for velocity?

Chemistry
1 answer:
laiz [17]3 years ago
7 0
The correct answer is meters/seconds
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The electron configuration of an element is 1s2 2s2. How many valence electrons does the element have?
faust18 [17]

Answer:

Two valence electrons

Explanation:

7 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
How many g of Fe2O3 will be produced from 132.7 g of o2
zhuklara [117]

Answer:

73.46839716589713698731965

Explanation:

4 0
3 years ago
Which of the following will produce a precipitate?
nekit [7.7K]

Reaction of option c produces precipitate.

Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.

Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.  

Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:

Co^2^+ + Zn(NO¬_3 )¬_2    Co(NO¬_3 )¬_2    + Zn^2^+

Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution  


8 0
3 years ago
Help fast !
jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0
3 years ago
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