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vfiekz [6]
3 years ago
15

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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Which is the correct choice
umka21 [38]
Answer:2 Fr because is has the lowest electronegativity 0.79
4 0
2 years ago
Could you please Calculate the number of atom of 40K (potassium 40) in 1gram of KCl. Taking into account the isotopic abundance
Crank

Answer:

9.53*10^{17} atoms of 40K

Explanation:

You can use the molecular mass and the Avogadro´s number, in the following formula:

N_{40K=\frac{m_{KCl}}{M_{MKCl}}}*N_{Avogadro}*(IA_{40k})

where m_{KCl} is the sample mass, M_{KCl} is the molecular mass of the KCl and IA(40K) is the isotopic abundance of 40K.

Now replacing the values, you can find:

N_{40K}=\frac{1g}{74.5513\frac{g}{mol}}*6.022*10^{23}mol^{-1} *0.000118

N_{40K}=9.53*10^{17} atoms

8 0
3 years ago
Helllppppppppp....<br>don't guess !!!​
noname [10]

Answer:

it answer D because waste is flowing thru the air

7 0
3 years ago
Read 2 more answers
MY NOTES ASK YOUR TEACHER Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of N
Alekssandra [29.7K]

Answer:

Two moles of HC2H3O2 react with one mole of Ca(OH)2 to produce one mole of calcium acetate and two moles of water.

Explanation:

HC2H3O2 is Acetic acid that can also be represented as (CH3COOH).

when Ca(OH)2 reacts with Acetic acid the product formed will be Calcium acetate and water

Chemically the reaction can be represented as

2CH 3 COOH + Ca(OH) 2    →      Ca(CH 3 COO) 2   +    2H 2 O

Two moles of CH3COOH react with one mole of Ca(OH)2 to produce one mole of Ca(CH3COO)2 and two moles of H2O.

7 0
3 years ago
Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r
Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid,  the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄  NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).  

5 0
3 years ago
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