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Finger [1]
3 years ago
14

Why aren't there any plate boundaries in Maryland???

Chemistry
1 answer:
algol [13]3 years ago
3 0
Because their are no plate boundaries near or next to maryland

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How many molecules are in 3.50 moles of H2O<br>​
Debora [2.8K]

Answer:

2.11 x 10²⁴ molecules.

Explanation:

  • <em>It is known that every 1.0 mole of a molecule contains Avogadro's number of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:</u></em>

1.0 mole of H₂O contains → 6.022 x 10²³ molecules.

3.5 mole of H₂O contains → ??? molecules.

∴ 3.5 mole of H₂O contain = (3.5 mol)(6.022 x 10²³) = 2.11 x 10²⁴ molecules.

5 0
3 years ago
If an acid has a ka=1.6x10^-10, what is the acidity of the solution?
LUCKY_DIMON [66]
The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
 In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+.  If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).
5 0
3 years ago
The chemical equation for a reaction between K2Cr2O7 and HCl is shown. K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O
guapka [62]

<u>Answer: </u>The correct answer is Option b.

<u>Explanation:</u>

Oxidizing agent is defined as the chemical reagent which helps the other chemical compound to get oxidized and itself gets reduced. The oxidation state for these species gets reduced because they are undergoing reduction reaction.

For the given chemical equation:

K_2Cr_2O_7+14HCl\rightarrow 2CrCl_3+2KCl+3Cl_2+7H_2O

Oxidation state of Chromium is getting reduced from +6 to +3 and oxidation state of chlorine getting increased from -1 to 0.

Hence, K_2Cr_2O_7 acts like and oxidizing agent because it is itself getting reduced to CrCl_3

Therefore, the correct answer is Option b.

4 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
Which of the following is an example of observational scientific investigation?
Softa [21]
B: nothing the color of the solution when heated
8 0
3 years ago
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