All categories

3 years ago

Whoever can answer this I will give all of my points to and brainliest award.

Chemistry

2 answers:

ZanzabumX [31]3 years ago

7 0

i dont see the picture.

Serga [27]3 years ago

4 0

The picture doesn’t pop up

You might be interested in

If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79

Helen [10]

1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass.

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

$200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4$

3) convert moles of CuSO₄ to moles of Cu

$1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu$

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

$1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu$

I did it step-by-step as the explanation but you can do all of this in one step.

$200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu$

4 0

3 years ago

What is not indicated by a reaction rate?

inna [77]

The answer is C. The speed at which Calories are consumed is not indicated by a reaction rate, rather it is indicated by what you do in a day.

3 0

3 years ago

How many moles are there in 45.0 grams of sulfuric acid,<br> H2SO4?

olga2289 [7]

Answer:

the answer is 0.4588162459

Explanation:

1 mole = 0.010195916576195

= 45 ×"

5 0

2 years ago

Plz help me with this question, I am stuck

zhenek [66]

i only know that E is one of them

7 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago