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Troyanec [42]
2 years ago
12

A particle with 10 protons, 11 neutrons, and 12 electrons has a mass number of:

Chemistry
1 answer:
garik1379 [7]2 years ago
7 0

For any given element the mass number can be found by adding the protons and the neutrons.

In this case its 10 protons plus 11 neutrons which gives us 21 as the mass number.

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Imagine that A and B are cations and X, Y, and Z are anions, and that the following reactions occur:
Gekata [30.6K]
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
</span>
AX(aq) is soluble and <span>BY(aq) is insoluble

the answer is
</span><span>E. BY</span>
5 0
3 years ago
An object has a mass of 3.50 grams and a density of 5.61 g/mL. What is the volume of this object? *
Elina [12.6K]

Answer:

<h2>0.62 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{3.50}{5.61}  \\  = 0.6238859...

We have the final answer as

<h3>0.62 mL</h3>

Hope this helps you

5 0
2 years ago
Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
7 0
2 years ago
What is the name of the compound with the formula PbF2?
Ber [7]

Lead fluoride hope this helps

4 0
3 years ago
Help please !!!!!!!!
Elena L [17]

Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

5 0
3 years ago
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