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3 years ago

What is called when a paragraph studies data carefully

Chemistry

2 answers:

Zigmanuir [339]3 years ago

8 0

Paraphrasing and summarizing

Vadim26 [7]3 years ago

4 0

✅Paraphrasing and summarizing .✅

IamSugarBee

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State two causes of cracked and sore nipples during breastfeeding

Nadusha1986 [10]

The biting and sucking of the baby when eating.

5 0

3 years ago

Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the follo

kiruha [24]

Explanation:

The given reaction equation will be as follows.

$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

Let is assume that at equilibrium the concentrations of given species are as follows.

$[Fe^{3+}] = 8.17 \times 10^{-3}$ M

$[SCN^{-}] = 8.60 \times 10^{-3}$ M

$[FeSCN^{2+}] = 6.25 \times 10^{-2}$ M

Now, first calculate the value of $K_{eq}$ as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

= $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

= $11.24 \times 10^{-4}$

Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

7 0

3 years ago

What are the reactants in a neutralization reaction?

castortr0y [4]

answer an element and a compound

8 0

3 years ago

A sample of 87.6 g of carbon is reacted with 136 g of

Vadim26 [7]

Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon = 87.7g

Mass of fluorine gas = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?

Solution:

Equation of the reaction:

C + 2F₂ → CF₄

let us find the number of the moles the given species;

Number of moles = $\frac{mass}{molar mass}$

C; molar mass = 12;

Number of moles = $\frac{87.7}{12}$ = 7.31moles

F; molar mass = 2(19) = 38g/mol

Number of moles = $\frac{136}{38}$ = 3.58moles

So;

From the give reaction:

1 mole of C requires 2 moles of F₂

7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

So;

2 moles of F₂ will produce mole of CF₄

3.58 moles of F₂ will then produce $\frac{3.58}{2}$ = 1.79moles of CF₄

6 0

3 years ago

If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0

3 years ago