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bearhunter [10]

3 years ago

15

Why are scientific theories changed but never thrown away?

2 answers:

Lesechka [4]3 years ago

5 0

Scientific theories provide results that are based on evidences that comes from research. These theories provide laws. But with more research sometimes certain results that comes from theories found not to be valid and some flaws are found that required to be removed. So that the theory required to be changed but only that theory provide the basis for the further research so it is never thrown away.

cluponka [151]3 years ago

4 0

Because the theories might be true and they should never be thrown away

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An amide may be produced by reacting an acid chloride with

Elanso [62]

Answer:

An amide may be produced by reacting an acid chloride with ammonia.

8 0

4 years ago

Read 2 more answers

What was the scientific basis for the unit amu

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Natural abundance of oxygen I think

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3 years ago

Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium

Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

5 0

4 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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4 years ago

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Answer:

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3 years ago