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faust18 [17]
2 years ago
13

Fill in the blanks and balance with the correct coefficient​

Chemistry
1 answer:
MrRissso [65]2 years ago
4 0

\huge\mathrm{  \underline{Answer}}࿐

The Balanced equation will be :

\mathrm{ \boxed2Fe +  \boxed3Cl_2  \rightarrow  \boxed2FeCl_3}

The Coefficients are :

  • Fe - 2

  • Cl_2 - 3

  • FeCl_3 - 2

_____________________________

\mathrm{ \#TeeNForeveR}

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A saturated solution can become supersaturated under which of the following conditions
Kaylis [27]

the answer is d. when more solute is added

5 0
2 years ago
How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94
erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is 16.23\times 10^{-5}g.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = 380\times 10^{-3}      (1ml=10^{-3}L)

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

pH+pOH=14\\pOH=14-8.94=5.06

pOH=-log[OH^-]

5.06=-log[OH^-]

[OH^-]=0.00000871=8.71\times 10^{-6}mole/L

Now we have to calculate the moles of OH^-.

Formula used : Moles=Concentration\times Volume

\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles

For neutralization, equal number of moles of H^+ ions will neutralize same number of OH^- ions.

\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles

As, H_2SO_4\rightarrow 2H^++SO^{2-}_4

From this reaction, we conclude that

2 moles of H^+ ion is given by the 1 mole of H_2SO_4

3309.8\times 10^{-9} moles of H^+ ion is given by \frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9} moles of H_2SO_4

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of H_2SO_4 × Molar mass of sulfuric acid

Mass of sulfuric acid = (1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g

Therefore, the mass of sulfuric acid needed is 16.23\times 10^{-5}g.

3 0
3 years ago
Which of these statements explains the relationship among elements, compounds, and mixtures? A. Elements are made of compounds a
sweet-ann [11.9K]
The answer is answer D
5 0
3 years ago
Read 2 more answers
Calculate the pH value in each of the following solutions, given their [H3O+] concentrations.
FrozenT [24]

Answer:

See Explanations ...

Explanation:

In general, pH is a 'p-factor' expression which, simply put, is a way to express very small numbers (i.e.; exponential data with 10⁻ⁿ value ranges) in a more convenient form. That is, by definition, pX = -log(X) where X is the data value of interest. In practical terms, p-factor analysis can be applied to a number of physical & chemical measurements such as ...

pH => measure of acidity of solution = -log[H₃O⁺]

pOH => measure of alkalinity of solution = -log[OH⁻]

pKa => measure of weak acid ionization in aqueous solution = -log(Ka)

pKb => measure of weak base ionization in aqueous solution = -log(Kb)

pKsp => measure of salt ionization in aqueous solution = -log(Ksp)

Such can be applied to ranges of small-number values defining other chemical and physical properties.

For this problem:

Gastric Juice: [H₃O⁺] = 1.6 x 10⁻²M => pH = -log(1.6 x 10⁻²) = -(-1.80) = 1.8

Cow's Milk:  [H₃O⁺] = 2.5 x 10⁻⁷M => pH = -log(2.5 x 10⁻⁷) = -(-6.60) = 6.60

Tomato Juice:  [H₃O⁺] = 5.0 x 10⁻⁵M => pH = -log(5.0 x 10⁻⁵) = -(-4.30) = 4.30

Other Applications:

Given:

[OH⁻] = 6.30 x 10⁻¹³M => pOH = -log(6.30 x 10⁻¹³) = -(-12.2) = 12.2

Ka = 4.5 x 10⁻⁵ => pKa = -log(4.5 x 10⁻⁵) = -(-4.35) = 4.35

Kb = 8.2 x 10⁻⁶ => pKb = -log(8.2 x 10⁻⁶) = -(-5.09) = 5.09

Ksp = 5.5 x 10⁻¹⁰ => pKsp = -log(-5.5 x 10⁻¹⁰) = -(-9.26) = 9.26

Note: The values for Ka, Kb & Ksp are typically provided in tables of weak acid ionization constants (Ka-values), weak base ionization constants (Kb-values) or solubility product constants of salts (Ksp-values).

Hope this helps, Doc :-)

4 0
2 years ago
Which evidence did Alfred Wegener use as evidence for the Continental Drift Theory?
Digiron [165]
For this problem
D all of the above
6 0
3 years ago
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