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MAVERICK [17]
2 years ago
15

The wavelength of the characteristic, bright yellow-orange flame test color of sodium is 590 nm. Calculate the average energy (Δ

E) associated with this atomic emission line.
Chemistry
1 answer:
Assoli18 [71]2 years ago
7 0
<span>The energy associated with the 590nm atomic emission line is 3.369 x 10^49 Joules. This can be determined by multiplying Plank’s constant (6.626 x 10^34 Js) by the speed of light (3x 10^8 m/s), and then dividing the result by the wavelength (590 x 10^-9 m).</span>
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A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat
Free_Kalibri [48]

Answer:

-154KJ/mol

Explanation:

mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter

= 0.2 × 100 / 1000  ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.

Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K

where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK

heat gain by calorimeter  = (4.184 × 200 + 480) × 2.34 = 3081.3 J

ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol

8 0
2 years ago
What mass of carbon dioxide could be made from 100 tonnes of calcium carbonate?
MA_775_DIABLO [31]
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3




i.e 100 tonnes of CaCO3 .


mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2




mass % of CO2 in CaCO3 = (44/100)×100 =44%


i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.

therefore, 44% of CO2 is present in CaCO3.








3 0
2 years ago
For a solar eclipse to occur which of the following alignments is necessary? A. The moon is located along a straight line betwee
svlad2 [7]

A. I'm pretty sure, could be C though..

8 0
2 years ago
2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp
weeeeeb [17]

Answer : The grams of N_2 consumed is, 89.6 grams.

Solution : Given,

Mass of CaCN_2 = 265 g

Molar mass of CaCN_2 = 80 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate the moles of CaCN_2.

\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles

The given balanced reaction is,

CaC_2+N_2\rightarrow CaCN_2+C

from the reaction, we conclude that

As, 1 mole of CaCN_2 produces from 1 mole of N_2

So, 3.2 moles of CaCN_2 produces from 3.2 moles of N_2

Now we have to calculate the mass of N_2

\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2

\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g

Therefore, the grams of N_2 consumed is, 89.6 grams.

5 0
2 years ago
Read 2 more answers
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