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lianna [129]
2 years ago
3

A large sample of solid calcium sulfate is crushed into smaller pieces for testing. which two physical properties are the same f

or both the large sample and one of the smaller pieces?
Chemistry
2 answers:
kiruha [24]2 years ago
5 0
Solubility and density I believe 
Svetllana [295]2 years ago
4 0

Explanation:

A property that does not bring any change in chemical composition of a substance are known as physical properties.

For example, change in shape, size, mass, volume, density, solubility etc are all physical properties.

So, when we crush a large sample of solid calcium sulfate into smaller pieces for testing then their density and solubility will remain the same. But their size and shape will definitely change.

As density is the mass divided by volume. Total mass of solid calcium sulfate is not changed during the testing. Hence, it will remain the same in a given volume.

Thus, we can conclude that two physical properties are the same for both the large sample and one of the smaller pieces is that their solubility and density.

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How do moss leaves and fish differ? How are they the same?
Lapatulllka [165]
Moss leaves and fish are different in that, the moss leave is a producer, that is, it produces its own food through photosynthesis while the fish is a consumer, it feeds on foods that are not produced by it.  
Both moss and fish are the same in the sense that both have cell as their basic unit of life, that is, they both possess cells.
4 0
2 years ago
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak
Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0
2 years ago
Which of the following is a valid conversion factor?
VMariaS [17]

Answer:

100 cg/1g

Step-by-step explanation:

    1 cg = 0.01 g     Multiply by 100

100 cg = 1 g

(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.

(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.

(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.

7 0
2 years ago
Read 2 more answers
Candace is finding the boiling point of water. From her research, she knows the boiling point of pure water is 212°F. When she r
Anestetic [448]

I would say that Candace's answer is d. wide-ranging. she didn't get the exact / precise (they mean the same thing) answer.

6 0
2 years ago
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Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio
Paha777 [63]
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

6 0
2 years ago
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