How can a flame test identify an unknown element?

Chemistry

1 answer:

Ilya [14]3 years ago

6 0

Answer:

the answer for this question is the option D

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If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?

GenaCL600 [577]

K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent

KNO₃ is limiting reagent

5 0

3 years ago

Given the following balanced equation, determine the rate of reaction with respect to [SO2].

Gnoma [55]

Answer:

Rate = -1/2 Δ[SO<sub>2</sub>]/Δt

so its gonna be (in more simple terms) rate= -1/2Δ(SO2)/Δt

Explanation:

7 0

2 years ago

(20 POINTS!!) please help me write a short reference page on chemical bonds (anything helps)

Shtirlitz [24]

Explanation:An ionic bond essentially donates an electron to the other atom participating in the bond, while electrons in a covalent bond are shared equally between the atoms. The only pure covalent bonds occur between identical atoms. ... Ionic bonds form between a metal and a nonmetal. Covalent bonds form between two nonmetals.

7 0

2 years ago

Which conversion factor is needed to solve the following problem?

USPshnik [31]

Molar Volume is required to solve this problem. As we know that "1 mole of any gas at standard temperature and pressure occupies 22.4 L of volume". SO using this concept, we can calculate the volume of ammonia formed by reacting 54.1 L of Hydrogen gas as follow,

3 0

3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago