A student goes into a lab and performs an experiment which produces 4.56grams

One way to measure temperature in some applications is to monitor the gas pressure in a rigid, closed container. What is the tem

julia-pushkina [17]

Answer:

296.43 K is the temperature of a vessel when pressure reads 1.250 atm.

Explanation:

At STP, when pressure is 1.000 atm the value of temperature is 273,15 K.

If the pressure at temperature T reads as 1.250 atm.

$P_1=1.000 atm, T_1=298.15K$

$P_2=1.250 atm,T_2=?$

Applying Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$ (At constant volume)

$\frac{1.000 atm}{273.15 K}=\frac{1.250 atm}{T_2}$

$T_2=296.43 K$

296.43 K is the temperature of a vessel when pressure reads 1.250 atm.

6 0

3 years ago

HELP! PLEASE! ASAP! Manganese(III) fluoride, MnF3, can be prepared by the following reaction:

Crank

Answer:

14.336 g MnF₂

Explanation:

number of moles = mass / molecular weight

number of moles of MnI₂ = 55 / 309 = 0.178 moles

number of moles of F₂ = 55 / 38 = 1.447 moles

From the reaction and the number of moles calculated we deduce that the fluorine F₂ is a limiting reactant.

So:

if 13 moles of F₂ reacts to produce 2 moles of MnF₃

then 1.447 moles of F₂ reacts to produce X moles of MnF₃

X = (1.447 × 2) / 13 = 0.223 moles of MnF₃ (100% yield)

For 57.2% yield we have:

number of moles of MnF₃ = (57.2 / 100) × 0.223 = 0.128 moles

mass = number of moles × molecular weight

mass of MnF₃ = 0.128 × 112 = 14.336 g

3 0

3 years ago

Calculate the mass (in grams) of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard, assumi

horrorfan [7]

1 part per million is the same as

1 mcg/mL (mcg = 1 micro gram = 1 * 10^-6 grams). 
125 ppm = 125 mcg/mL 

125 mcg/ml * [1 gram / 1 * 10^6 mgrm] = 1.25 * 10^-4 grams 

The density of hard water is 1.000 grams / mL 

50 mL of water contains 1.25 * 10^-4 grams. 

given mass = 1.25 * 10^-4 
Molar mass = 100.0 grams / mole 
n = ????? 

n = 1.25 * 10^-4/100 = 1.25 * 10^-6 moles.

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7 0

3 years ago

The entropy of a system at 337 K increases by 221.7 J/mol•K. The free energy value is found to be –717.5 kJ/mol. Calculate the c

OlgaM077 [116]

Answer: The change in enthalpy for the given system is -642.8 kJ/mol

Explanation:

To calculate the change in enthalpy for given Gibbs free energy, we use the equation:

$\Delta G=\Delta H-T\Delta S$