Which are always the first waves of an earthquake to be detected

When completing an experiment, results will be much more accurate if multiple trials are completed and the results averaged. TRU

grandymaker [24]

When completing an experiment, results will be much more accurate if multiple trials are completed and the results averaged. True. That's because you may get different results each time.

5 0

3 years ago

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A scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. How many grams of na3po4

natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

2 years ago

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago

One main difference between a Prokaryotic cell and an Eukaryotic cells is __________.

sasho [114]

The Eukaryotic cell has a nucleus

8 0

2 years ago

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How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of

Drupady [299]

For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.

5 0

3 years ago

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