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artcher [175]
3 years ago
7

Plz help me with this question plzzzzz​

Chemistry
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

6

Explanation:

Formula: Al2O3

If we require 2Al2O3

                       We divide 2 by 3

Sorry for the shadow of phone and fingers.

 

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
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Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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F belongs to Group 7A in the periodic table so it has 7 valence electrons. Each F needs to share 1 pair of electrons to complete the octet.

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Answer:

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3) Molar mass of Li_{2}SO_{4} = 7x2 + 32 + 16x2 = 78g

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