All categories

3 years ago

What substance has a melting point of 0 and a boiling point of 100

Chemistry

2 answers:

tiny-mole [99]3 years ago

7 0

Answer:

Water

Explanation:

Water molecules are known to have a melting point of 0°C and a boiling point of 100°C.

The melting and boiling points are as a result of the intermolecular force of attraction between the molecules of water. The predominant bond type that dictates the property of the melting and boiling point is the hydrogen bonding. This bond is a strong force of attraction and it impacts on the physical properties of water.

nika2105 [10]3 years ago

7 0

Answer:

Water

Explanation:

The substance that has a melting point at 0 and a boiling point at 100 is water. Water is the only substance that fits into this category. According to van.physics.illinois.edu, it states the following "For pure water, the boiling point is 100 degrees Celsius (212 Fahrenheit) at one atmosphere of pressure, and the melting point is 0 degrees Celsius (32 degrees Fahrenheit) at one atmosphere of pressure."

Answer: Water

You might be interested in

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

PLS HELP Students in a chemistry class added 5 g of Zinc (Zn) to 50 g of hydrochloric acid (HCl). A chemical reaction occurred t

Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is 5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

is 0.154 g. Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =

10.42 g

4 0

3 years ago

The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer

sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

$pH=pKa+log(\frac{salt}{acid} )$

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0

3 years ago

Given that Kp [NOTE: Kp!!!!] = 1.39 at 400 ºC for the reaction, P4(g) <=> 2 P2(g), which answer best describes the reactio

Igoryamba

Answer:

The reaction will proceed to the left to attain equilibrium.

Explanation:

The question is missing but I guess it must be about how the reaction will proceed to attain equilibrium.

First, we have to calculate the partial pressures using the ideal gas equation.

$pP_{4}=\frac{2.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L} =5.52atm$

$pP_{2}=\frac{1.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L}=3.31atm$

Now, we have to calculate the reaction quotient (Qp).

$Qp=\frac{pP_{2}^{2}}{pP_{4}} =\frac{3.31^{2} }{5.52} =1.98$

Since Qp > Kp, the reaction will proceed to the left to attain equilibrium.

3 0

3 years ago

Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma

Elena-2011 [213]

Answer:

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

Explanation:

$CaCO_3(s)\rightarrow CaO(s) + CO_2(g)$

Annual production of CaO = $8.6\times 10^{10} kg=8.6\times 10^{13} g$

Moles of CaO :

$\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles$

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with $1.53\times 10^{12} moles$ of CaO moles of carbon-dioxide moles produced will be:

$\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles$ of carbon-dioxide

Mass of $1.53\times 10^{12}$ moles of carbon-dioxide:

$1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg$

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

6 0

3 years ago