I need help asap why is there only one variable in an experiment that changes?

1 answer:

5 0

Answer: If more than one variable is changed in an experiment, scientist cannot attribute the changes or differences in the results to one cause. By looking at and changing one variable at a time, the results can be directly attributed to the independent variable.

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How does burning affect an object's property? Burning can cause an object to increase in volume. Burning changes the chemical ma

Jobisdone [24]

Answer: Burning changes the chemical make up of an object.

Explanation:

A chemical change can be defined as a change in the substance when it combines with other kind of substance to form a new substance. A chemical change can also occur when a substance is broken down into two or more products. These changes cannot be reversed. These changes affect the physical make up of an object. For example, burning as when an object is burned it cannot be transformed into its original form. A wood if burned can be converted into ash, water and carbon dioxide cannot regain its original form after burning so burning brings about chemical change in an object.

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3 years ago

What characteristic of an element’s atoms always determines the element’s identity?

Anna11 [10]

Answer:

hporntue dhdjehwgs r. rvegdyfuee

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3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$