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svetlana [45]
3 years ago
6

Select all that apply. A scientific law: A- is generally accepted as true B- has never been found to be in error C- has no known

contradictions D- must be reverified occasionally
Chemistry
2 answers:
Anon25 [30]3 years ago
6 0

A and D

They are sometimes proven wrong and contradicted, so these would make the most sense

Sidana [21]3 years ago
4 0

Answer:

A) is generally accepted as true

D)must be re verified occasionally

Explanation:

A scientific law is proposed based on some scientific studies including hypothesis, thesis , verification etc.

These when proposed are considered as true that is why they are accepted.

However with time they must be re verified occasionally so that if chance of some contradiction and error can be minimized. For example periodic law proposed by Mendleev was based on atomic weight which was later on changed to be based on atomic number.

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A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

3 0
3 years ago
Which gas would act most like an ideal gas? a. chlorine gas at 32oC b. chlorine gas at -185oC
Maslowich
A as the lower temperature in Celsius corresponds to a lower Kelvin temperature thus reducing movement
6 0
3 years ago
9.0 mol Na2S can from 9.0 mol CuS and 8.0 mol CuSO4 can form 8.0 mol Cus.
ICE Princess25 [194]

Answer:

765.0 grams CuS

Explanation:

The limiting reagent is the reactant which completely reacts before the other reactant(s) is used up. When 9.0 moles Na₂S and 8.0 moles CuSO₄ react, it appears that CuSO₄ is the limiting reagent. You can tell because it results in the production of less product.

You can determine the mass of CuS by multiplying the moles by the molar mass. It is important to arrange the ratio in a way that allows for the cancellation of units.

Molar Mass (CuS): 95.62 g/mol

8.0 moles CuS               95.62 g
-------------------------  x  -----------------------  =  765.0 grams CuS
                                         1 mole

4 0
2 years ago
According to Bernoulli's Principle, slower moving fluids exert ____________ pressure than faster moving fluids. (Fill in the bla
Mariana [72]
According to Bernoulli's Principle, slower moving fluids exert a lower or decreased pressure than faster moving liquids. 
7 0
3 years ago
A 3.24-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio
Stolb23 [73]

Answer:

a) mass of the H2CO3 produced:

given:

Mass of sample = 3.24 g

Mass of Na2CO3 obtained after decomposition = 2.19 g

Solution :

Molar mass of NaHCO3 = 84

reaction:

2NaHCO3 → Na2CO3 + H2CO3

so it is clear that 2 mole of NaHCO3 gives 1 mole of Na2CO3 and H2CO3

Now, ICE table for the reaction is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x +x +x

E 3.24/84 -2x x x

As NaHCO3 is completely decomposed so final Concentration of NaHCO3 is zero.

=> 3.24/84 -2x = 0

=> 2x = 3.24/84

=> x = 1.62/84

The new ICE table is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x = -2(1.62/84) +x = 1.62/84 +x = 1.62/84

E 0 1.62/84 1.62/84

From the above ICE table,

it is found that (1.62/84 ) moles of H2CO3 is obtained.

Since,

The molar mass of H2CO3 is 62

=> Mass of H2CO3 obtained = moles × molar mass

=> Mass of H2CO3 obtained = (1.62 /84 ) × 62

= 1.19 grams

Mass of H2CO3 experimentally :

Mass of reactants = mass of products

=> Mass of sample = mass of Na2CO3 + mass of H2CO3

=> Mass of H2CO3 = mass of sample - mass of Na2CO3

= 3.24 - 2.19 = 1.05 g

b) Experimental mass = 1.05 g

Theoretical mass = 1.19 g

Percentage yield of H2CO3 = Experimental mass × 100 / Theoretical mass

= 1.05 × 100 /1.19

= 88.23 %

6 0
3 years ago
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