87,036 mi Express your answer as an integer.

What elements are in NaC2HO4 and how many atoms are in each element

JulijaS [17]

Answer:

Each molecule contains one atom of A and one atom of B. The reaction does not use all of the atoms to form compounds.

A + B ⟶ Product

Particles: 6 8 6

If six A atoms form six product molecules, each molecule can contain only one A atom.

The formula of the product is ABₙ.

If n = 1, we need six atoms of B.

If n = 2, we need 12 atoms of B. However, we have only eight atoms of B, so the formula of the product must be AB.

Thus, 6A + 6B ⟶ 6AB, with two B atoms left over.

Explanation:

Credit goes to @znk

Hope it helps you :))

7 0

2 years ago

Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu

Nitella [24]

Answer:

$\Delta \:H_{vap}=40383.88\ J/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 13.95 torr

$P_2$ = 144.78 torr

$T_1$ = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

$T_1$ = 298.15 K

$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

4 0

2 years ago

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan

Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0

2 years ago

4.8g of calcium is added to 3.6g of water. The following reaction occurs

notka56 [123]

Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present

q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present

sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present

8 0

2 years ago

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

Aleksandr-060686 [28]

To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s

5 0

2 years ago

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