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Ainat [17]
3 years ago
9

The answer please thank you

Chemistry
1 answer:
otez555 [7]3 years ago
5 0

Answer:

B

Explanation:

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What discovery was made about particles with an accelerator in 1998
kramer

Answer/Explanation:

In June 1998 in Japan a scientist discovered that neutrinos (which is a type of particle) has weight, mass. This was later proven with some very convincing strong evidence.

<u><em>~ LadyBrain</em></u>

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3 years ago
which characteristics of an element can be determined by considering only the element’s specific location on the periodic table?
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Gas, Solid, or liquid

Explanation:

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Read 2 more answers
Write chemical equation for the following reactions. Indicate physical states and write net ionic equations. If no reaction indi
iren [92.7K]

There is no reaction.

<em>Molecular equation :</em>

K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)

<em>Ionic equation :</em>

2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)

<em>Net ionic equation :</em>

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>

<em>All ions cancel</em>. There is no net ionic equation.

7 0
3 years ago
2,65 Lt de un gas ideal se encuentran a 25°C y 1,2 atm de presión. Si se pasa el gas a un nuevo recipiente de 1,5 litros a una t
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Answer:

2.5 atm

Explanation:

P2 = 1.2 atm x 2.65 L x 345 K / 1.5 L x 298 K

Esto debería ser igual...

2.45 atm (personajes importantes)

4 0
3 years ago
Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final
Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

<u>Step 1:</u> Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

<u>Step 2:</u> The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

<u>Step 3:</u> Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

<u>Step 4: </u>Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

<u>Step 5:</u> Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

<u>Step 6:</u> Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0
3 years ago
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