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3 years ago

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

1 answer:

8 0

Molecular equation:
$H_{2}SO_{4}(aq) + 2NaOH(aq)--\ \textgreater \ Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Complete ionic equation:

2H^{+}+SO_{4}^{2-} + 2Na^{+} +2OH^{-}--\ \textgreater \ 2Na^{+}+SO_{4}^{2-} + 2H_{2}O(l)


Near 2H^{+},SO_{4}^{2-} , 2Na^{+} , OH^{-} 

should be written (aq).

Remove ions that do not change from both sides

2H^{+} + 2OH^{-}=2H_{2}O

Net ionic equation:

H^{+}(aq) + OH^{-}(aq)=H_{2}O(l)

Answer is D.

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Pure water and pure salt are poor conductors of electricity. When salt is dissolved in water, the resulting solution conducts el

castortr0y [4]

Answer:

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Explanation:

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2 years ago

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lakkis [162]

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2 years ago

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How many grams of HF are needed to react with 3.0 moles of Sn?

Flauer [41]

Answer:

120g

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction.

Sn + 2HF —> SnF2 + H2

Step 2:

Determination of the number of mole HF needed to react with 3 moles of Sn.

From the balanced equation above,

1 mole of Sn and reacted with 2 moles of HF.

Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.

Step 3:

Conversion of 6 moles of HF to grams.

Number of mole HF = 6 moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn.

3 0

2 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

Significant figures are an indicator of the certainty in measurements true or false?

IRISSAK [1]

True.

SF are used for simplifying figures in a measurement to produce a more accurate reading.

3 0

3 years ago