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2 years ago

Please help me, Thank you!

Chemistry

2 answers:

weeeeeb [17]2 years ago

8 0

Answer:

the second one

Explanation:

im positive bc ur gaining 2 ions

Usimov [2.4K]2 years ago

4 0

The second one because the little two beside the o is what you are gaining ;)

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I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP

GalinKa [24]

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

3 0

2 years ago

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

sashaice [31]

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm

Ideal gas equation: pV = nRT => n = pV / RT

R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm

=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol

2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type

X (+) + O2 (g) ---> X2O or

2 X(2+) + O2(g) ----> X2O2 = 2XO or

4X(3+) + 3O2(g) ---> 2X2O3

In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)

So, lets probe those 3 cases.

3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X

Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol

That does not correspond to any of the metal with valence 1+

So, now probe the case 2.

4) Case 2:

2moles X metal / 1 mol O2(g) = x / 0.01226 mol

=> x = 2 * 0.01226 = 0.02452 mol

And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol

That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.

4) Case 3

4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol

That does not correspond to any metal.

Conclusion: the identity of the metallic element could be titanium.

5 0

3 years ago

I need to find the orbit notation<br> 1. Ne - 1s 2s 2p 3s 3p 4s 3d 4p 5s

weeeeeb [17]

Answer:

1s*2 2s*2 2p*6

Explanation:

atomic number of neon is 10 so the s-shell contains 2 electron and p-shell contains 6

3 0

2 years ago

snow_lady [41]

The suggestion is to prevent a puddle of the liquid present in the sample from forming or from it leaking on to the surface on which it is placed. For example, if precipitates of a solid are removed from water and then placed on filter paper to dry, the water will soak into the filter paper and then leak on to the counter on which it is placed. If this precipitate were placed in a watch glass or weighing paper, the water would only evaporate and would not contaminate the sample.

5 0

3 years ago

If element A is in Group 13 and element B is in Group 17, which ions will be formed when A and B react together?

Kobotan [32]

Answer:

A3+ and B-

Explanation:

Elements in group 13 have outermost electron configuration, ns2np1 hence they form trivalent positive ions.

Elements in group 17 have outermost electron configuration ns2np5 hence they form univalent negative ions.

This implies that, if element A is in Group 13 and element B is in Group 17, the ions formed are A3+ and B-.

8 0

2 years ago