Evidence used to support or refute a scientific theory or hypothesis is known as material evidence.
Evidence is also used by scientists in various contexts, such as when they apply ideas to real-world issues.
Whether a person accepts the data as proof depends on their presumptions or views regarding how observations relate to a theory.
A creative hypothesis is developed by a scientist and may be refuted by testing it against evidence or established truths.
Evidence can refute a hypothesis by demonstrating facts that are at odds with it. In contrast, the presence of evidence does not preclude the possibility of future, undiscovered evidence supporting the idea.

Therefore, it is the power of material evidence that wins, not the authority of experts that

proves an answer in Science.

Learn more about hypothesis here: brainly.com/question/606806

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7 0

2 years ago

Given the unbalanced equation: PbCl₂ + Na₂CrO₂ ---> PbCrO₂ + NaCl. When properly balanced, the sum of the balancing coefficie

Veronika [31]

Answer:

A. 2

hope this helps :)

if you need an explanation let me know!!

3 0

3 years ago

What is the molarity if 2.00 liters containing 49.0 grams of sodium carbonate [Na2CO3)?

yKpoI14uk [10]

Answer: The molarity of solution is 0.231 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Molar mass of $Na_2CO_3$ = $2\times 22.99+1\times 12.01+3\times 16.00=105.99$

moles of $Na_2CO_3$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{49.0g}{105.99g/mol}=0.462mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.462mol}{2.00L}$

$Molarity=0.231M$

Therefore, the molarity of solution is 0.231 M

5 0

3 years ago

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

3 years ago