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pentagon [3]
2 years ago
11

The density of mercury is 13.6 g/cm3 . What volume (in quarts) is occupied by 100. g of Hg? (1 L = 1.06 qt)

Chemistry
1 answer:
SpyIntel [72]2 years ago
6 0

Answer:

0.00077 qt

Explanation:

Density -

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass mercury = 100 g

Density of mercury = 13.6 g/cm³ .

Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -

d = m / V

13.6 g/cm³ = 100 g  / V

V = 7.35 cm³

1 cm³ = 0.001 L

V = 7.35 * 0.001 L = 0.0073 L

Since ,

1 L = 1.06 qt

V = 0.0073* 1.06 qt = 0.0077 qt

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The correct answer(reported to the proper number of significant figures)to the following is
LenKa [72]
to the proper number of significant figures) to the following? (12.67+19.2)(3.99)/(1.36+ 11.366).
8 0
1 year ago
How many atoms of zirconium are in 0.3521 mol of zirconium?
lora16 [44]

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

3 0
2 years ago
How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?
Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

Moles of NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles  

W_s = weight of the solvent in g = ?

0.452=\frac{0.348\times 1000}{W_s}

W_s=770g

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

5 0
1 year ago
After recrystallization, the final weight of biphenyl is found to be 0.02 g. Assume that all steps were carried out correctly, n
Lubov Fominskaja [6]

Explanation:

From the source, The student recrystallized biphenyl by using benzene. Biphenyl is non polar and so is benzene. Biphenyl is very soluble in benzene at low as well as high temperature. Thus, it do not recrystallize as like dissolves like and gives a very poor yield.

The good recovery of the product is dependent on compound having high solubility at the high temperatures and having low solubility at the low temperatures.

6 0
2 years ago
In each row, check the box under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g. have t
vredina [299]

Answer:

HCH_{3}SO_{2}

H_{3}PO_{3}

HClO_{2}

Explanation:

Every acid (HA) tends to disolve into proton (H^{+}) and anion (A^{-}) in aqueous solution. Acid strength can be determined by measuring this tendency to separate into proton an anion. Strength of an acid can be quantified by its acid dissociation value - Ka. A strong acid will have a tendency to easily release proton and will have larger Ka value and smaller logarithmic value (pKa = - logKa) similar to calculating pH of the solution. So the easiest way to resolve this issue is by looking for Ka or pKa value of the acid (This table may be useful in more complex tasks and is attached below). However, stronger acid can be determined elsehow.

a) Carbon is element 14 with 4 valent electrons and sulfur is element 16 with 6 valence electrons. Thus, sulfur has stronger electronegativity (tendency to attract bonded electrons towards itself). This means that sulfur will hold oxygen tighter to itself so the hydrogen bond to it can be more easily separated from it. HCH_{3}SO_{2} is more acidic in aqueous solution.

b) In H_{3}PO_{4}, phosphorus holds one double bond with oxygen and three OH group equally. To show an acidic tendency, phosphorus would need to let go one hydrogen out of one of OH groups. In H_{3}PO_{3}, phosporus holds two double bong with oxygen, one OH and one hydrogen, all single and lonely, ready to leave phosphorus and show acidic characteristics in aqueous solution. Thus, H_{3}PO_{3} is more acidic compound.

C) In all Cl acids, the electron density is placed around Cl so the more oxygen around Cl, the more acidic will be the chemical. This is comparable to an oxidation state - the bigger oxidation state, the stronger acid will be:

HClO_{4}  ^{+7} >HClO_{3}^{+5} >HClO_{2}^{+3} >HClO_{}^{+1}

HClO_{2} can reasonably be expected to be more acidic in aqueous solution.

4 0
2 years ago
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