Complete the balanced equation: Cu(NO3)2 + Na2CO3

A gas is held under conditions of standard temperature and pressure. It is found that 44.0 grams of the gas occupies a volume of

storchak [24]

Answer:

CO2

Explanation:

(I Just took the test)

At STP 1 mol=22.4 Liters, so we now know that it is asking for which of the gasses has a molar mass of 44, and CO2 is th only one with that molar mass

5 0

3 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.

7 0

3 years ago

A cube has a mass of 42 grams and a volume of 15 cubic centimeters. What is it’s density?

deff fn [24]

Answer:

2.8g/cm³

Explanation:

Given parameters:

Mass of cube = 42g

Volume of cube = 15cm³

Unknown:

Density of the cube = ?

Solution:

Density is defined as the mass per unit volume of a substance. It is mathematically expressed as:

Density = $\frac{mass}{volume}$

So;

Density = $\frac{42}{15}$ = 2.8g/cm³

6 0

3 years ago

A student prepares a solution by combining 100 mL of 0.30 M HNO2(aq) and 100 mL of 0.30 M KNO2(a). Which of the following equati

katrin2010 [14]

Answer:

B

Explanation:

This creates a buffer solution meaning it will just neutralize out and does not change the PH massively.

7 0

3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago

Complete the balanced equation: Cu(NO3)2 + Na2CO3 > ?