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3 years ago

How many moles are in 275g (Ca(NO3)2

Chemistry

2 answers:

shepuryov [24]3 years ago

4 0

Answer:

1.68 mol Ca(NO₃)₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Compounds

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 275 g Ca(NO₃)₂

[Solve] mol Ca(NO₃)₂

Step 2: Identify Conversions

[PT] Molar Mass of Ca: 40.08 g/mol

[PT] Molar Mass of N: 14.01 g/mol

[PT] Molar Mass of O: 16.00 g/mol

Molar Mass of Ca(NO₃)₂: 40.08 + 2[14.01 + 3(16.00)] = 164.10 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 275 \ g \ Ca(NO_3)_2(\frac{1 \ mol \ Ca(NO_3)_2}{164.10 \ g \ Ca(NO_3)_2})$
[DA] Divide [Cancel out units]: $\displaystyle 1.67581 \ mol \ Ca(NO_3)_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.67581 mol Ca(NO₃)₂ ≈ 1.68 mol Ca(NO₃)₂

Anvisha [2.4K]3 years ago

4 0

Answer:

1.68 mol Ca(NO₃)₂

Explanation:

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A sample of helium occupies a volume of 180.0 ml at a pressure of 0.800 atm and a temperature of 29°C. What will be the

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Answer:

If the volume of the container is decreased to 90.0 ml and the pressure is increased to 1.60 atm the temperature will be 302 °K

Explanation:

Boyle's law says that "The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure", which is expressed mathematically as:

P * V = k

Charles's Law consists of the relationship between the volume and temperature of a certain amount of ideal gas, which is maintained at a constant pressure, by means of a proportionality constant that is applied directly. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} =k$

Finally, Gay Lussac's law establishes that as the temperature increases, the gas molecules move more quickly and therefore the number of collisions against the walls increases, that is, the pressure increases since the container is of fixed walls and its volume can not change. Therefore, the ratio of pressure to temperature always has the same value (it is constant).

$\frac{P}{T}=k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T}=k$

Having a state 1 at the beginning and a state 2 at the end is fulfilled:

$\frac{P1*V1}{T1}=\frac{P2*V2}{T2}$

In this case, you know:

P1= 0.800 atm
V1= 180 mL= 0.180 L (being 1 L=1000 mL)
T1= 29 °C= 302 °K
P2= 1.60 atm
V2= 90 mL= 0.09 L
T2=?

Replacing:

$\frac{0.800 atm*0.180L}{302 K}=\frac{1.60 atm*0.09L}{T2}$

Solving:

$T2*\frac{0.800 atm*0.180L}{302 K}=1.60 atm*0.09L$

$T2=\frac{1.60 atm*0.09L}{\frac{0.800 atm*0.180L}{302 K}}$

T2= 302 °K

If the volume of the container is decreased to 90.0 ml and the pressure is increased to 1.60 atm the temperature will be 302 °K

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3 years ago

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3 years ago

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3 years ago

CaCO3(s) + HCl(aq) -> CO2(g) + H2O(l) + CaCl2(aq)

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3 years ago

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at 58.3°C, the

Elan Coil [88]

Answer:

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

$T_{a}$ = 58.3°C

$L1_{a}$ = 747mmHg

$L2_{a}$ = 52mmHg

$T_{b}$ = 110°C

$L1_{a}$ = 577mmHg

$L2_{b}$ = 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

$T_{a}$ = 58.3°C + 273.2

$T_{a}$ = 331.5k

$T_{b}$ = 110°C + 273.2

$T_{b}$ = 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

$P^{0}$ = $P_{atm} - (P_1 -P_2)$

The vapor pressure under the first set of conditions is:

$P^{a}$ = 755mmHg - (747mmHg - 52mmHg)

$P^{a}$ = 60mmHg

The vapor pressure under the second set of conditions is:

$P^{b}$ = 755mmHg - (577mmHg = 222mmHg)

$P^{b}$ = 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

$In_p^{0} = \frac{- (delta) H}{RT} + B$

where (delta) H = ΔH

The slope of the $In_p^{0} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

Calculating the slope; we have:

$\frac{- (delta) H}{R} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

$\frac{- (delta) H}{R}$ = $\frac{331.5 * 383.2 * In (400/60)}{(331.5-383.2)}$

$\frac{- (delta) H}{R}$ = -4661.4k

$\frac{ (delta) H}{R}$ = 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B = $In_p_{1} + \frac{ (delta) H}{RT}$

B = $In 60 + \frac{4661.4}{331.5}$

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P = $\frac{-4661.4k}{T} + 18.156$

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

$In_p_{0}$ = $In_p_{0} = \frac{-4661.4}{403.2} + 18.156$

$In_p_{0}$ = 6.595

$p_{0}$ = $e^{6.595}$

$p_{0}$ = 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

$P_1(130^{0}C)$ = 731.44mmHg

$P_1(58.3^{0}C)$ = 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

$y_1 = \frac{P^0 (T)}{P}$