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2 years ago

3. Mercury is a liquid metal that has a density of 13.58 g/mL. Calculate the volume of mercury

Chemistry

1 answer:

Alona [7]2 years ago

5 0

Answer:

0.03682 mL of mercury

Explanation:

We know the density of the mercury which is 13.58 g/mL

density = mass / volume

volume = mass / density

Now we can calculate the volume of 0.5 g of mercury:

volume = 0.5 / 13.58 = 0.03682 mL of mercury

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A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevatio

Goryan [66]

Answer:

MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

$\Delta T=imK_b\qquad\text{where i=ions/mole, m=molality, $K_b\approx 0.512$}\\\\376.81-373.15=(x+1)\dfrac{\text{moles}}{\text{kg solvent}}(0.512)\\\\\dfrac{3.66}{0.512}=(x+1)\dfrac{\dfrac{30.2}{56+35.45x}}{0.1}=\dfrac{302(x+1)}{56+35.45x}\\\\\dfrac{3.66}{0.512\cdot 302}(56+35.45x)=x+1\\\\\dfrac{3.66\cdot 56}{0.512\cdot 302}-1=x\left(1-\dfrac{3.66\cdot 35.45}{0.512\cdot 302}\right)\\\\x=\dfrac{50.336}{24.877}\approx 2.023$

Then the formula for the salt is MCl₂.

6 0

2 years ago

Which of the reagents listed below would efficiently accomplish the transformation of ethyl-3-pentenoate into 3-penten-1-ol?

Mandarinka [93]

Reactions of Ethyl-3-pentenoate with all given reagents are given below.

Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.

Reaction with NaBH₄:
Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.

Reaction with LiAlH₄:
Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.

Result:
The correct answer is Option-A (Highlighted RED below).

7 0

2 years ago

Predict what will observe in below mention experiment.

Sloan [31]

Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature

a. It is likely that more rock candy will be formed in batch A.

b. It is likely that less rock candy will be formed in batch A.

c. It is likely that no rock candy will be formed in either batch.

d. I need more information to predict which batch is more likely to form rock candy.

Answer: Option A

Explanation:

More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.

Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.

The sugar will be dissolved in water until the time all the space is filled sugar molecules.

Hence, the correct answer is Option A.

6 0

2 years ago

Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta

Arlecino [84]

Answer:

A. $\lambda_0=2.196\times 10^{-7}\ m$

Explanation:

The work function of the Platinum = $9.05\times 10^{-19}\ J$

For maximum wavelength, the light must have energy equal to the work function. So,

$\psi _0=\frac {h\times c}{\lambda_0}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda_0$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

Thus,

$9.05\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\frac{9.05}{10^{19}}=\frac{19.878}{10^{26}\lambda_0}$

$9.05\times \:10^{26}\lambda_0=1.9878\times 10^{20}$

$\lambda_0=2.196\times 10^{-7}\ m$

8 0

2 years ago

If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Arada [10]

Find moles of MgSO4.7H2O

molar mass = 246
so moles = 32 / 246 = 0.13 moles.

When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g

so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g

6 0

2 years ago