Whats the molar mass of tetraphosphorus decoxide

What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H

Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

5 0

2 years ago

Mrs. Johnson's science class perform an experiment. They combine two substances that react with lots of fizzing and bubbling. Th

s2008m [1.1K]

Chris is correct because the reactants and products do not have to have the same mass, but they do have to weigh the same. Is the correct answer:) Hannah is right because the mass of the reactants was different than the mass of the products. is incorrect

3 0

3 years ago

Read 2 more answers

How much energy is needed to heat 2.00g of carbon from 50.0 degrees Celsius to 80.0 degrees Celsius If carbon specific heat capa

Sedaia [141]

Answer:

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Explanation:

7 0

3 years ago

How does a sample of water at 38 °C compare to a sample of water at 295 K?

Sidana [21]

Converting the temperature, 295 K from Kelvin to Celsius scale:

$295 K - 273 = 22^{0} C$

Water has a boiling point of $100^{0}C$ and a melting point of $0^{0} C$

When we compare water at two different temperatures, $22^{0}C and 38^{0}C$ we can say that water is in liquid form at both these temperatures as both of them are quite below the boiling temperature and above the melting temperature.

The temperature difference between water at the given two temperatures = $38^{0}C - 22^{0}C = 16^{0}C$

Water at $38^{0}C$ is at a higher temperature and so is warmer than water at a lower temperature of $22^{0}C (or 295 K)$ .

7 0

3 years ago

Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8

kaheart [24]

Answer : The average rate of consumption of $H^+$ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$