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Write a balanced overall reaction from these unbalanced half-reactions sn and ag

loris [4]

Balance each one by adding electrons to make the charges on both sides the same:

Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag

Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:

2 Ag+ + 2 e- --> 2 Ag

Now, just add the two half reactions together, cancelling anything that's the same on both sides:

2 Ag+ + Sn --> Sn2+ + 2 Ag

And you're done.

8 0

3 years ago

Identifying Qualitative and quantitative Data

lyudmila [28]

Answer:

qualitative data : quantitative data :

circular in shape 75 colonies ...

stained purple 200 purple ..

spreads across plate 55 colonies ...

Explanation:

i got it right :) .

7 0

3 years ago

The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum m

topjm [15]

Answer:

<u><em>1.7 × 10³ kg of ore.</em></u>

Explanation:

Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.

Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:

X × 71% × 82% = 1.0 × 10³ kg

↑ ↑ ↑ ↑

(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)

You must just simplify, solve and compute:

0.71 × 0.82 × X = 1,000

X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg

Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.

6 0

3 years ago

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Hydrochloric acid reacts with calcium to form hydrogen and calcium chloride. if 100 grams of hydrochloric acid reacts with 100 g

natita [175]

The balanced chemical reaction is:

2HCl + Ca = CaCl2 + H2

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

100 g HCl ( 1 mol HCl / 36.46 g HCl ) = 2.74 mol HCl

100 g Ca ( 1 mol Ca / 40.08 g ) = 2.08 mol Ca

From the reaction, the mole ratio of the reactants is 2:1 where every 2 moles of hydrochloric acid, 1 mole of calcium is required. Therefore, the limiting reactant for this case is calcium.

4 0

3 years ago

Read 2 more answers

The answer to question 2

Bad White [126]

Answer:

12

Explanation:

You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.

$M_{r}$ : 258.21 18.02

KAl(SO₄)₂·xH₂O ⟶ KAl(SO₄)₂ + xH₂O

Mass/g: 4.74 2.16

Step 1. Calculate the mass of the KAl(SO₄)₂.

Mass = 4.74 g – 2.16 g = 2.58 g.

Step 2. Calculate the moles of each product.

$\text{Moles of KAl(SO}_{4})_{2} = \text{2.58 g} \times \frac{\text{1 mol} }{\text{258.21 g}} = 9.992 \times 10^{-3} \text{ mol}$

$\text{Moles of H}_{2}\text{O} = \text{2.16 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 0.1200 mol}$

Step 3. Calculate the molar ratio of the two products.

$\frac{\text{Moles of KAl(SO}_{4})_{2}}{\text{Moles of H}_{2}\text{O}} = \frac{ 9.992 \times 10^{-3} \text{ mol}}{\text{ 0.1200 mol} } = \frac{1 }{12.01} \approx \frac{ 1}{ 12}$

1 mol of KAl(SO₄)₂ combines with 12 mol H₂O, so x = 12.

3 0

4 years ago