Ask question

Ask question

All categories

3 years ago

8

A sample of carbon monoxide gas is initially in a 5858 mL container. The gas is then moved to 3.29 L container at a temperature

of 195 °C. Determine the initial temperature of the gas in kelvin. Round your answer to three significant figures. Include the unit in your answer.

1 answer:

Doss [256]3 years ago

5 0

Answer:

d i took the test

Explanation:

You might be interested in

If a piece of metal were to cut in half would the density change

nadezda [96]

Density depends on both the mass and the volume of an object. If you cut a bar of gold in half, you would have two bars with half the mass of the original bar. However, each bar would also have half the volume of the original bar. The density of gold does not change.

8 0

3 years ago

Read 2 more answers

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

How many grams of S are in 475 g of SO2?

enot [183]

There are 237. 5 g of Sulfur,S in 475 g of SO2?

<h3 />

<h3>Calculation of grams of Sulfur</h3>

From the question, we can say that

The molar mass of sulfur = 32 g/mol

The molar mass of oxygen = 16 g/mol

Therefore,

The molar mass for SO2 = 32 + (16 × 2) g/mol = 64 g/mol

Now,

If 1 mole of SO2 contains 1 mole of S

Then 64 g of SO2, will contain 32g of S;

Such that

475 g of SO2 will give { $\frac{ 32g of S) ( 475 g of SO2)}{64 of SO2}$ } = 237. 5 g of Sulfur.

Learn more about molar mass here :brainly.com/question/18291695

6 0

2 years ago

A substance that reduces the hydrogen ion concentration of a solution is __________.

ioda

A substance that reduces the hydrogen ion concentration of a solution is <u>base</u>.

A base which thus completely dissociates in an aqueous solution is referred to as a strong base. These substances produce one or more hydroxide ions (OH-) per base molecule when they ionize in water. A weak base, on the other hand, only partially splits into its water-soluble ions.

The three definitions of "base" used in chemistry are Arrhenius base, Bronsted base, as well as Lewis base. The fact instead that bases react towards acids is acknowledged by all base definitions.

Therefore, a substance that reduces the hydrogen ion concentration of a solution is <u>base</u>.

To know more about hydrogen ion

brainly.com/question/5327882

#SPJ4

8 0

2 years ago

Which describe beta decay? Check all that apply 1.In beta decay, a proton becomes a neutron and a positron 2.In beta decay, a ne

ivanzaharov [21]

haha thanks again for the help.

3 0

3 years ago

Read 2 more answers