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3 years ago

Does pebbles dissolve in water?

Chemistry

2 answers:

Lyrx [107]3 years ago

6 0

Pebbles of salt or regular pebbles?
If it's regular pebbles than no

PolarNik [594]3 years ago

3 0

Hi there!

The answer is no.

Although the real answer depends on what the pebble is composed of (since different materials produce different effects), most pebbles don't dissolve in water.

Hope this helps!

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Can you Guys please help me on 14,and 15

jok3333 [9.3K]

14. is a and i think 15. is b

7 0

2 years ago

Read 2 more answers

[04.02]What coefficients must be added to balance the following equation?

vesna_86 [32]

Methink this is a neutralization reaction where an acid reacts with a base to form salt and water; but if we must balance the equation, we need to know what the product would be? So our base LiOH reacts with our acid Tetraoxosulphate (VI) H2SO4. The reaction produces salt and water as evidence. LiOH + H2SO4 gives Li2SO4 and H20. We need to make sure that the total atoms on the LHS and RHS balanced. So adding two moles of LiOH, we have 2LiOH + H2SO4 produces LiSO4 + H20. The eqn isn't balanced yet as there's 2 moles of deficit H2 atoms on the RHS, So our final reaction we have 2LiOH + H2SO4 gives Li2SO4 + 2H2O. Hence our answer is C

7 0

3 years ago

Recovery standards are a necessary tool for determining exactly how much of a particular analyte you are able to extract from a

Dafna11 [192]

Answer:

2.05mg Fe/ g sample

Explanation:

In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.

In the problem you recover 3.5mg Fe / 1.0101g sample: 3.465mg Fe / g sample. As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:

3.465 / 4×100 = 86,6%

As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:

1.7mg Fe / 0.9582g sample× (100/86.6) = 2.05mg Fe / g sample

I hope it helps!

5 0

3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

Read 2 more answers

One day on her walk home from school, Leann noticed some rust on an old car.

skad [1K]

Answer:

Iron and Oxygen

Explanation:

Fe is Iron and O2 is Oxygen

4 0

2 years ago