Ask question

Ask question

All categories

gizmo_the_mogwai [7]

2 years ago

9

How can the existence of spectra help to prove that energy levels in atoms exist?

1 answer:

Liula [17]2 years ago

6 0

The existence of spectra can help the existence of atoms by expanding and multiplying in the law of London Dispersion

You might be interested in

A teacher wants to figure out The amount of a solution that is needed for a titration experiment in a class lab! the class has 1

AnnZ [28]

The teacher needs 1500 mL for the whole class.

5 0

2 years ago

Read 2 more answers

How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7

JulsSmile [24]

Ok so this is what we know :

2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0

2 years ago

Read 2 more answers

When 1.365 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the

luda_lava [24]

<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Heat released by anthracene= Heat absorbed by water

Heat absorbed by water =

$\tt Q=500\times 4.18\times 25.89=54110.1~J$

mol of anthracene (MW=178,23 g/mol)

$\tt \dfrac{1.365}{178.23}=0.00766$

The enthalpy of combustion per mole of anthracene :

$\tt \Delta H=-\dfrac{Q}{n}=\dfrac{54110.1}{0.00766}=-7063981.7~J/mol\approx -7064~kJ/mol$

8 0

1 year ago

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so

pickupchik [31]

Answer:

C. 4x10⁻⁴ mol / (Ls)

Explanation:

Based in the reaction:

5 H₂O₂(aq) + 2 MnO₄⁻(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 O₂(g)

2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = <em>4x10⁻⁴ mol / (Ls)</em>

Right answer is:

C. 4x10⁻⁴ mol / (Ls)

8 0

2 years ago

The molar absorptivity of a compound at 500 nm wavelength is 252 M-1cm-1. Suppose one prepares a solution by dissolving 0.00140

adell [148]

Answer:

The absorbance of the solution is 0.21168.

Explanation:

Given that,

Wavelength = 500 nm

Molar absorptivity = 252 M⁻¹ cm⁻¹

Number of moles = 0.00140

Volume of solution = 500.0 mL

Length = 3.00 mm

We need to calculate the molar concentration

Using formula of the molar concentration

$C=\dfrac{N}{V}$

Where, N = number of moles

V = volume

Put the value into the formula

$C=\dfrac{0.00140}{0.5000}$

$C=0.0028\ M$

We need to calculate the absorbance of the solution

Using formula of absorbance

$A=\epsilon C l$

Put the value into the formula

$A=252\times0.0028\times0.300$

$A=0.21168$

Hence, The absorbance of the solution is 0.21168.

5 0

2 years ago