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Reil [10]
3 years ago
15

A mercury lamp emits radiation with a frequency of 5.62 x1024 Hz. Calculate the wavelength of

Chemistry
1 answer:
Nitella [24]3 years ago
7 0

Answer:

wavelength = 0.534×10⁻¹⁶ m

Explanation:

Given data:

Frequency of wave = 5.62 ×10²⁴ Hz

Wavelength = ?

Solution:

Speed of photon = wavelength × frequency

wavelength = speed of photon / frequency

Now we will put the values in formula:

wavelength = 3×10⁸ m/s / 5.62 ×10²⁴ Hz

Hz = s⁻¹

wavelength = 3×10⁸ m/s / 5.62 ×10²⁴ s⁻¹

wavelength = 0.534×10⁻¹⁶ m

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The second reaction in the formation of sulfuric acid occurs slowly.
rjkz [21]

<u>Answer:</u> The correct option is NO_2(g)

<u>Explanation:</u>

A catalyst is defined as the chemical species that increases the reaction rate but does not participate in it and is left behind after the completion.

A homogeneous catalyst is one that is present in the same phase as the reactants and products.

A heterogeneous catalyst is one that is present in different phase as that of reactants and products.

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

As all the reactants and products are in gaseous state so, the homogeneous catalyst must also be in the gaseous state only.

Hence, the correct option is NO_2(g)

5 0
3 years ago
In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is
V125BC [204]

Answer:

the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

Explanation:

Given that:

mass of diphenylacetylene (C_{14}H_{10}) = 0.5297 g

Molar Mass of diphenylacetylene (C_{14}H_{10}) = 178.21 g/mol

Then number of moles of diphenylacetylene (C_{14}H_{10})  = \frac{mass}{molar \ mass}

= \frac{0.5297  \ g }{178.24 \  g/mol}

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = Heat absorbed by H_2O + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene (C_{14}H_{10}) will be = \frac{20598.87 \ J}{0.002972 \ mol}

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

3 0
3 years ago
Read 2 more answers
Compute 214.056 + 9.3456. Round the answer appropriately
shusha [124]
214.0560
+ 9.3456
=223.4016
3 0
3 years ago
Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is
Pani-rosa [81]
<span>294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
6 0
3 years ago
Compounds can not be separated by Response area means. Separating a compound requires a Response areareaction.
Aleks04 [339]

Answer:

use vinager

Explanation:

I did this heh :/

7 0
3 years ago
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