The gross colors can be imparted to the flame by the metal ion solutions .
1.To serve as an excitation source, turn on a Bunsen burner.
2. Perform flame tests on the chloride solutions of Li+, Na+, K+, Ca2+, Sr2+, and Ba2+ as instructed by your lab instructor. Wearing the unique glass blower glasses, take note of the obscene hue that each ion contributes to the flame. With one exception: in order to view the sodium flame, you must remove the special glasses that suppress sodium emissions.
3. Obtain two unidentified answers and note their numerical values. Using the ugly hue the solution gives the flame as a guide, identify the metal ions that are present.
One of the six ions you tested will make up the particle.
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Answer:
NA + F2 = NAF - Chemical Equation Balancer.
2 Na + F 2 → 2 NaF
Na is a reducing agent, F 2 is an oxidizing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.
Answer:
Explanation:
BrO3- (aq) + Sb^3+ (aq) --------> Br^3- (aq) + Sb^5+ (aq) is an unbalanced equation and needs to be balanced
BrO3- (aq) → Br^3- (aq
to balance it water must be added to the right side and H⁺ be added to the left side
BrO₃⁻ + 6 H⁺ + 8e⁻ → Br³⁻ + 3 H₂ O
Sb³⁺ (aq) → Sb⁵⁺ + 2e⁻
multiply the second equation by 4
BrO₃⁻ + 6 H⁺ + 8e⁻ → Br³⁻ + 3 H₂ O
4Sb³⁺ → 4Sb⁵⁺ + 8 e⁻
add the two equation together and cancel the 8 e electrons on both side
BrO₃⁻ + 4Sb³⁺ + 6 H⁺ → Br³⁻ + 4Sb⁵⁺ + 3 H₂ O
number of mole of BrO₃⁻ = volume in liters × molarity = (29.9 / 1000) L × 0.120 M = 0.003588 moles
from the balanced equation of reaction;
one mole of BrO₃⁻ requires 4 moles of Sb³⁺
0.003588 moles of BrO₃⁻ will require = 0.003588 × 4 = 0.0144 moles of Sb³⁺
a) amount of antimony in grams in the sample = 0.0144 moles × 121.8 g ( molar mass of antimony) = 1.748 g
b ) percentage of antimony in the ore = 1.748 g / 6.33 g = 27.62 %
Answer:
2.0 l
Explanation:
When for 20 °C volume is 1 l then for 40°C is going to be 2 l
20-------1
40-------x
x=(40*1):20