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PSYCHO15rus [73]
3 years ago
5

Why is the sky blue and not red, even though blue light scatters more easily than red.

Chemistry
1 answer:
balu736 [363]3 years ago
6 0
Because molecules in the air scatter blue light from<span> the sun </span>more than<span> they </span>scatter red light<span>.</span>
You might be interested in
Question 14 of 17
artcher [175]

Since Lutetium-177 is a beta and gamma emitter, the daughter nuclide produced from the decay of this radioisotope is 177Hf.

Beta emission of a radioisotope yields a  daughter nuclide whose amass number is the same as that of its parent nucleus but its atomic number is greater is greater than that of the parent nucleus by 1 unit.

Also, gamma emission does not lead to any change in the mass number of atomic number of the daughter nucleus produced.

Hence, the stable daughter nuclide, 177Hf is produced.

Learn more: brainly.com/question/1770619

4 0
3 years ago
You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution
babymother [125]

Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

7 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
When writing the chemical formulas for a molecular compound, what method do you use?​
podryga [215]

Answer:

A molecular compound is usually composed of two or more nonmetal elements. Molecular compounds are named with the first element first and then the second element by using the stem of the element name plus the suffix -ide.

4 0
3 years ago
How much heat do you need to raise the temperature of 100g of aluminum from 30 C to 150 C
SVEN [57.7K]

Answer:

Q = 10.8 KJ

Explanation:

Given data:

Mass of Al= 100g

Initial temperature = 30°C

Final temperature = 150°C

Heat required = ?

Solution:

Specific heat of Al = 0.90 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30°C

ΔT = 120°C

Q = 100g×0.90 J/g.°C× 120°C

Q = 10800 J       (10800j×1KJ/1000 j)

Q = 10.8 KJ

4 0
3 years ago
Read 2 more answers
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