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3 years ago

Sort these elements into pairs that would most likely exhibit similar chemical properties.

1 answer:

3 0

The pair of elements that would most likely exhibit similar chemical properties are:
K and Li belong to Group 1
Te and S belong to Group 6
Kr and Ne belong to Group 8
Elements under the same group have similar chemical properties.

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How are tadpoles and larvae similer

Bond [772]

Answer: Tadpole, also called polliwog, aquatic larval stage of frogs and toads. Compared with the larvae of salamanders, tadpoles have short, oval bodies, with broad tails, small mouths, and no external gills. The internal gills are concealed by a covering known as an operculum.

Explanation:

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3 years ago

Solve the following problem. Give your answer to the correct number of significant figures. a. 20.0 meters x 0012.65 meters b. 0

Nezavi [6.7K]

Explanation:

For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count.

a. 20.0 meters x 0012.65 meters

The numbers are;

20.0 (3 s.f) and 12.65 (4 s.f)

The multiplication gives; 253

Since 253 is already in 3 s.f, that's the answer.

b. 002.5 × 103 meters + 3.50 × 102 meters

The numbers are;

002.5 (2 s.f), 103 (3 s.f), 3.50 (3 s.f) and 102 (3 s.f)

002.5 × 103 = 257.5 = 260 (2 s.f)

3.50 × 102 = 357 = 360 (2 s.f)

260 + 360 = 620

4 0

3 years ago

The top is an example problem the bottom one is the one i need help with. I only need C,D, and E please help

velikii [3]

Answer:

correct

Explanation:

makes sense just re-write!

5 0

3 years ago

Read 2 more answers

How many milliliters of a 1:2000 drug "i" solution and a 7% drug "i" solution mixed would make 120ml of a 3.5% solution of drug

yanalaym [24]

To solve this problem, let us say that:

x = volume of 1:2000 drug "i" solution

y = volume of 7% drug "i" solution

Assuming volume additive, then this forms:

x + y = 120 mL

x = 120 – y ---> 1

1:2000 also refers to 0.0005 concentrations and 7% also refers to 0.07 concentrations. By doing a component balance:

0.0005 x + 0.07 y = 0.035 (120 mL)

0.0005 x + 0.07 y = 4.2

Substituting equation 1 into this derived equation to get an equation in terms of y:

0.0005 (120 – y) + 0.07 y = 4.2

0.06 – 0.0005 y + 0.07 y = 4.2

0.0695 y = 4.14

y = 59.568 mL = 59.57 mL

From equation 1, x would be:

x = 120 - 59.57

x = 60.43 mL

Answers:

59.57 mL of 1:2000 drug "i" solution

60.43 mL of 7% drug "i" solution

3 0

3 years ago

1.) Rust forms when Fe, O₂, and H₂O react in the balanced equation below;

valentinak56 [21]

Answer:

1.) A.) The limiting reactant is Fe.

B.) 16.17 g.

2.) 84.70 %.

Explanation:

For the balanced equation:

2Fe(s) + O₂(g) + 2H₂O(l) → 2Fe(OH)₂(s).

2.0 moles of Fe reacts with 1.0 mole of oxygen and 2.0 moles of water to produce 2.0 moles of Fe(OH)₂.

A.) Which of these reactants is the limiting reagent?

To determine the limiting reactant, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass.

Suppose that water is exist in excess.

no. of moles Fe = mass/atomic mass = (10.0 g)/(55.845 g/mol) = 0.179 mol ≅ 0.18 mol.

no. of moles of O₂ = mass/molar mass = (4.0 g)/(32.0 g/mol) = 0.125 mol.

Since from the balanced equation; every 2.0 moles of Fe reacts with 1.0 mole of oxygen.

So, 0.18 mol of Fe reacts with 0.09 mol of O₂.

Thus, the limiting reactant is Fe.

The reactant in excess is O₂ (0.125 mol - 0.09 mol = 0.035 mol).

B.) How many grams of Fe(OH)₂ are formed?

Using cross multiplication:

∵ 2.0 moles of Fe produce → 2.0 moles of Fe(OH)₂.

∴ 0.18 moles of Fe produce → 0.18 moles of Fe(OH)₂.

∴ The mass (no. of grams) of produced 0.18 mol of Fe(OH)₂ = no. of moles x molar mass = (0.18 mol)(89.86 g/mol) = 16.17 g.

2.) (Using the reaction listed in question 1.) If 2.00 g Fe is reacted with an excess of O₂ and H₂0, and a total of 2.74 g of Fe(OH)₂ is actually obtained, what is the % yield?

The % yield = [(actual mass/calculated mass)] x 100.

The actual mass = 2.74 g.

We need to calculate the theoretical mass:

Firstly, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass.

no. of moles Fe = mass/atomic mass = (2.0 g)/(55.845 g/mol) = 0.0358 mol ≅ 0.036 mol.

Using cross multiplication:

∵ 2.0 moles of Fe produce → 2.0 moles of Fe(OH)₂.

∴ 0.036 moles of Fe produce → 0.036 moles of Fe(OH)₂.

∴ The calculated mass (no. of grams) of produced 0.036 mol of Fe(OH)₂ = no. of moles x molar mass = (0.036 mol)(89.86 g/mol) = 3.235 g.

∴ The % yield = [(actual mass/calculated mass)] x 100 = [(2.74 g/3.235 g)] x 100 = 84.70 %.

4 0

4 years ago