Relate pressure to the area over which a force is applied.

(a) ions in a certain volume of .20M NaCl (aq) are represented in the box above on the left. in the box above on the right, draw

mamaluj [8]

Answer:

The first picture attached is the diagram that accompanies the question.

The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq) is:

CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0

2 years ago

Students in a science class learn that groundwater is an important source of drinking water for many communities. They construct

sveticcg [70]

Groundwater is also one of our most important sources of water for irrigation from septic tanks and toxic chemicals from underground storage tanks and leaky landfills ... Drinking contaminated groundwater can have serious health effects. If there is a leak, these contaminants can eventually make their way down

5 0

2 years ago

Water has a density of 1.0 g/ml. which of these objects will float in water? object i: mass = 50.0 g; volume = 40.2 ml object ii

Oksana_A [137]

<span>Object I would be it</span>

8 0

2 years ago

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

4 0

2 years ago

If we use 18 moles of H2 how many moles of H2O do we make? Use this chemical equation: 6H2 + O2 → 3H2O

Ilya [14]

<h3>Answer:</h3>

9 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6H₂ + O₂ → 3H₂O

[Given] 18 mol H₂

[Solve] mol H₂O

<u>Step 2: Identify Conversions</u>

[RxN] 6 mol H₂ → 3 mol H₂O

<u>Step 3: Stoich</u>

[DA] Set up conversion: $\displaystyle 18 \ mol \ H_2(\frac{3 \ mol \ H_2O}{6 \ mol \ H_2})$
[DA] Simplify: $\displaystyle 18 \ mol \ H_2(\frac{1 \ mol \ H_2O}{2 \ mol \ H_2})$
[DA] Divide [Cancel out units]: $\displaystyle 9 \ mol \ H_2O$

3 0

2 years ago

Read 2 more answers