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3 years ago

What causes an ice cube to melt when removed from a freezer?

Chemistry

2 answers:

OverLord2011 [107]3 years ago

4 0

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

iren2701 [21]3 years ago

3 0

The temperature in the room is higher than the temperature in the freezer, thus causing the melting state of the ice cube when it is removed from the freezer.

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What is the affect of this

Alisiya [41]

Answer:

Explanation:

Your answer should be in the attached pdf

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5 0

3 years ago

Read 2 more answers

What is the volume of an object with the mass of 7.9 grams in the density of 2.28g/ml.

Schach [20]

Answer:

3.7mL is the volume of the object

Explanation:

To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:

7.9g * (1mL / 2.28mL) =

<h3>3.7mL is the volume of the object</h3>

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3 years ago

TRUE FALSE? The MOLE is a unit in Chemistry that serves as a bridge between the ATOMIC and MACROSCOPIC worlds?

Vanyuwa [196]

Answer:

True

Explanation:

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

7 0

3 years ago

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What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?

inessss [21]

Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have
k = $\frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}$
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = $\frac{2.303}{18.9} X log \frac{100}{6.25}$ = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = $\frac{0.693}{k} = \frac{0.693}{0.1467}$ = 4.723 hours.

8 0

3 years ago

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

$\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}$

3 0

3 years ago