All categories

3 years ago

For which one of the following is ΔHfo zero?

Chemistry

2 answers:

Vlada [557]3 years ago

6 0

The answer is most likely C

Sholpan [36]3 years ago

3 0

The standard enthalpy of formation is zero for B and D since by definition, the standard enthalpy of formation for an element at its standard state is zero. The monatomic oxygen is not the standard state of oxygen while carbon dioxide is already a molecule.

You might be interested in

If 1.20 moles of an ideal gas occupy a volume of 18.2 l at a pressure of 1.80 atm, what is the temperature of the gas, in degree

scoundrel [369]

We can calculate for temperature by assuming the equation for ideal gas law:

P V = n R T

Where,

P = pressure = 1.80 atm

V = volume = 18.2 L

n = number of moles = 1.20 moles

R = gas constant = 0.08205746 L atm / mol K

Substituting to the given equation:

T = P V / n R

T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm / mol K)

T = 332.70 K

We can convert K unit to ˚C unit by subtracting 273.15 to Kelvin, therefore

T = 59.55 ˚C

6 0

4 years ago

What factors govern the position of an IR absorption peak? Select one or more correct answers.

Phoenix [80]

Answer:

The factors that govern the position of an IR absorption peak includes:

A.)strength of the bond

C.)masses of the atoms involved in the bond

D.)the type of vibration being observed

Explanation:

Infra red spectroscopy covers a range of techniques, mostly based on absorption spectroscopy.

4 0

3 years ago

In the reaction between Lithium Sulfate and an excess of Lead (II) Nitrate, how many molecules of Lithium Nitrate can be expecte

siniylev [52]

Answer:

$1.46x10^{23}molecules \ LiNO_3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Li_2SO_4 + Pb(NO_3)_2 \rightarrow PbSO_4+2LiNO_3$

Thus, since lead (II) nitrate is in excess, we can directly compute the moles of lithium nitrate by applying the 1:2 molar ratio between them in the chemical reaction as well as the molar mass of Lithium Sulfate that is 110 g/mol for the stoichiometric shown below factor:

$n_{LiNO_3}=13.3gLi_2SO_4*\frac{1molLi_2SO_4}{110gLi_2SO_4} *\frac{2molLiNO_3}{1molLi_2SO_4} =0.242molLiNO_3$

Finally, by using the Avogadro's number we are able to compute the molecules:

$0.242molLiNO_3*\frac{6.022x10^{23}molecules\ LiNO_3}{1mol}=1.46x10^{23}molecules \ LiNO_3$

Best regards.

5 0

4 years ago

Read 2 more answers

Name:

Liono4ka [1.6K]

Answer:

1. 505g is the mass of the aluminium.

2. The answer is in the explanation

Explanation:

1. To solve this question we need to find the volume of the rectangle. With the volume and density we can find the mass of the solid:

Volume = 7.45cm*4.78cm*5.25cm

Volume = 187cm³

Mass:

187cm³ * (2.702g/cm³) = 505g is the mass of the aluminium

2. When the temperature of a liquid increases, the volume increases doing the density decreases because density is inversely proportional to volume. And works in the same way for gases because the temperature produce more collisions and the increasing in volume.

4 0

3 years ago

A 4.10 −g sample of a mixture of CaO and BaO is placed in a 1.00-L vessel containing CO2 gas at a pressure of 735 torr and a tem

insens350 [35]

Answer:

The mass % of CaO is 9.37%%

Explanation:

Step 1: Data given

Mass of the mixture = 4.10 grams

Volume of the vessel = 1.00 L

The vessel contains CO2 with a pressure of 735 torr and temperature = 26 °C

The pressure of the remaining CO2 is 155 torr

Step 2: The balanced equation

CaO/BaO + CO2 → CaCO3 /BaCO3

Step 3: Calculate moles CO2 (via ideal gas law)

CO2 is the only gas, this means the drop in the pressure is caused by CO2

Δp = 735 torr - 155 torr = 580 torr

580 torr = 0.763 atm

n = PV/RT = (0.763atm)(1.00L) / (0.08206 L*atm/K*mol)(299) = 0.0391 moles CO2 initial

moles reacted = 0.0311 moles CO2

Step 4: Calculate mass of CaO and BaO

Let's suppose x = mass of CaO. Then mass of BaO = 4.10-x.

In the balanced equation, we notice the mole ratio of CaO/BaO with CO2 is 1:1. This means the number of moles CaO + moles BaO = Number of moles CO2

moles CaO + moles BaO = moles CO2

x/56.08 + (4.10-x)/153.33 = 0.0311

56.08 (x/56.08 + (4.10-x)/153.33) = (56.08)(0.0311)

x + (0.366)(4.10-x) = 1.744

x + 1.5006 - 0.366x = 1.744

0.634x = 0.2434

x = 0.384g CaO

4.1-x = g BaO = 3.716 grams

%CaO = (0.384 g / 4.10 g) x 100 = 9.366% ≈ 9.37

The mass % of CaO is 9.37%

7 0

4 years ago