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BlackZzzverrR [31]
3 years ago
13

The Density of this log is 1.75g/cm3. If we cut this log in half, what is the density of the log? Make sure to show your work an

d explain your answer.
Chemistry
1 answer:
swat323 years ago
5 0

Answer:

The density remains the same

Explanation:

The density remains the same because cutting the object in half will divide the mass & volume by the same amount. The density cant be changed no matter what happens to it.

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6 0
2 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
2 years ago
Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol
stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point):  K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI →  Na⁺  +  I⁻        i =2

b. CaCl₂ →  Ca²⁺  +  2Cl⁻      i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³     i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0
3 years ago
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