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tino4ka555 [31]
3 years ago
15

8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

kJ/mol. Determine the enthalpy of solution of RbF and indicate whether the solution temperature will increase or decrease when RbF is dissolved in water.
Chemistry
1 answer:
garri49 [273]3 years ago
3 0

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7)  kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

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How many moles of sodium cyanide are needed to react completely with 2.6 moles of sulfuric acid
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What is a solution of zinc and copper?<br>five letter word
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Answer:

Brass

Explanation:

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5 0
2 years ago
What is the mass of 0.55 mole<br> of magnesium chloride?
Gala2k [10]

Answer:

SYMBOLS, FORMULAS AND MOLAR MASSES

OBJECTIVES

1. To correctly write and interpret chemical formulas

2. To calculate molecular weights from chemical formulas

3. To calculate moles from grams using chemical formulas

INTRODUCTION

Part I. Symbols and formulas

An element is a homogeneous pure substance made up of identical atoms. All matter is made

up of elements and, since chemistry is the study of matter, it is convenient to use symbols to represent

the elements rather than using the entire name.

By international agreement, specific symbols are assigned to each element (Note: This means

that while names of the elements vary with language, symbols are constant throughout the world.) Each

element is assigned a one- or two-letter symbol. The first letter is capitalized, the second (if there is

one) is not. While this often seems trivial, it is in fact a very important point. For example, in chemical

language Co represents cobalt, which is a metal and an element, while CO represents carbon monoxide,

a compound which is a colorless, odorless gas! Even when there is not an obvious correspondence,

for instance "MN", it can cause confusion. Do you mean the element manganese? Did you forget a

letter and mean something else? Are you using "M" to represent something else entirely? Chemists

sometimes use "M" to represent any metal. It is well worth the trouble to memorize the symbols for

common elements.

Since compounds consist of elements, the chemical formulas of compounds also consist of

elements with subscripts used to denote the number of atoms per molecule. If there is no subscript, it is

implied that there is one of that kind of atom. Ones never appear in chemical formulas. Not only do

subscripts denote ratios of atoms, they also denote the ratio of moles of element to one mole of

compound. Parentheses can be used to show groups of atoms, with the subscripts showing how many

groups there are. Parentheses are not used if there is only one group.

Examples: For one mole of the following compounds, how many moles of each element are

present?

MgCl2 1 mole Mg, 2 moles Cl

Mg(NO3)2 1 mole Mg, 2 moles N, 6 moles O

NaNO3 1 mole Na, 1 mole N, 3 mole O

AgCl 1 mole Ag, 1 mole ClPart II. Molar Masses

Each atom has a different size and therefore a different mass. The relative masses of each

element can be found on the periodic table. For example, one atom of magnesium weighs 24.31 amu

(atomic mass units). However, one mole of magnesium weighs 24.31 g. (Moles were planned that

way!) Since one mole of MgCl2 consists of one mole of magnesium and two moles of chlorine, the

mass of one mole of MgCl2 must be the sum of the masses of the elements. The mass of one mole of a

substance is called the molar mass or molecular weight.

Examples: What is the molar mass of the following compounds?

MgCl2 24.31 + 2(35.45) = 95.21 g/mol

Mg(NO3)2 24.31 + 2(14.01) + 6(16.00) = 148.33 g/mol

NaNO3 23.00 + 14.01 + 3(16.00) = 85.01 g/mol

AgCl 107.9 + 35.45 = 143.4 g/mol

(Note: Yes! You DO have to count significant figures when calculating molecular weight/molar

mass. However, the number of significant figures may vary depending on which periodic table you use.)

Chemists are generally interested in number of moles. Unfortunately, it is impossible to measure

moles directly. However, masses are easily measured, and if the chemical formula of the compound is

known, the molar mass can be used to determine the number of moles. The molar mass is defined as:

molar mass = grams/moles = g/mol (1)

Moles may be calculated by using molar mass as a conversion factor in dimensional analysis where

molar mass in grams = 1 (exactly) mole of compound (2)

This method is used in multi-step calculations. For example, if 0.873 g of MgCl2 is weighed out, it

is 9.17 x 10-3

moles.

1 mole

0.873g x 95.21 g = 9.17 x 10-3

mol MgCl2 (3)

However, 0.873 g of AgCl is only 6.09 x 10-3

mol.

1 mole

0.873g x 143.4 g = 6.09 x 10-3

mol AgCl (4)Molar mass may also be used to relate moles to grams. For example, 0.158 mol of MgCl2 is 15.2 g.

0.158 mol x 95.21 g = 15.2 g MgCl2 (5)

1 mol

Percent is used to express parts per one hundred. Usually in chemistry, it refers to

g of species of interest x 100 = % (6)

g of whole thing

Example: For the % Mg in MgCl2: In one mole of MgCl2, there are 24.31 g of Mg (molar mass of Mg,

the part we are interested in) and 95.21 g of MgCl2 (the whole thing), so %Mg in MgCl2 is

(24.31/95.21) x 100 = 25.53% Mg (7)

PROCEDURE

Work individually.

The formula for calcium phosphate is Ca3(PO4)2. Weigh about 2 g of calcium phosphate to the

nearest 0.001 g. In other words, you do not have to have exactly 2.000g, but you must know the

weight you have exactly. Acceptable results include but are not limited to: 1.985g , 2.035g, 2.314g

etc.

Be sure to report all results with the correct number of significant figures and appropriate units!

5 0
3 years ago
A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

5 0
3 years ago
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