An explanation of what will happen every time in a specific situation

1 answer:

kramer3 years ago

4 0

The one that is being described above is what we call SCIENTIFIC LAW. Scientific law is what explains of what will happen every time in a certain situation. This is also different from scientific theory since scientific theory only gives an explanation to a group of happenings and this can still be modified. Hope this helps.

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PLEASE HELP QUICKLY! SERIOUS ANSWER PLEASE

yarga [219]

Answer:

Q1: A

Q2: B

Explanation:

Q1: the independent variable will always go on the x axis and the dependent will always go on the y-axis, so the answer is A.

Q2: it's not a, it's not d (ocean droughts aren't a thing that I'm aware of), and hurricanes don't make water scarce, so I would go with B.

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2 years ago

The two reactants combine to form a single product, ammonia, nh3, which is not shown. write a balanced chemical equation for the

nikdorinn [45]

If the product is ammonia, the reactants must be the diatomic gases N₂ and H₂. The complete balanced reaction is shown below:

<em>N₂ + 3 H₂ → 2 NH₃</em>

Now, the number of molecules on the left or reactant side are: 2 molecules of nitrogen and 3*3 = 6 molecules of hydrogen. <em>The total number would be 8 molecules. The same number of molecules is counted on the right or product side. </em>This is to obey the Law of Conservation of Mass.

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What is a hygrometer?

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A hygrometer is an instrument for measuring the humidity of the air or a gas .

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2 years ago

A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.

Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

$HBr+KOH\rightarrow KBr+H_2O$

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

$pOH=-log([OH^-])=-log(0.320)=0.50$

Thus, the pH is:

$pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50$

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

$n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol$

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

$n_{KOH}^{remaining}=0.00425mol$

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

$[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M$