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Agata [3.3K]
3 years ago
7

The isotope labeling an atom containing 40 protons and 51 neutrons suddenly has two neutrons added to it. What isotope is create

d.
Chemistry
1 answer:
marishachu [46]3 years ago
8 0

I believe the correct answer from the choices listed above is option 2. The isotope of an atom that contains 40 protons and 51 neutrons and suddenly has two neutrons added to it would be Zirconium-93. We look at a periodic table and find the element with the atomic number 40 which equal the number of protons.


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How salt solution can be determined by using hydrometer​
goldenfox [79]

Answer:

Salt solution may be calculated by measuring the specific gravity of a sample of water using a hydrometer.

<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>(</em><em>^</em><em>^</em><em>)</em><em>.</em><em>.</em><em>.</em><em>.</em>

3 0
4 years ago
17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI
galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: M_{1} = 0.20 M,      V_{1} = 15.0 mL

M_{2} = 0.10 M,            V_{2} = ?

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula s follows.

M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0
3 years ago
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,
Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

r^++r^-=\frac {a}{2}  ...................2

Given that:

Cl^-\ (r^-) = 1.82\ \dot{A}

To find,

K^+\ (r^+) = ? \dot{A}

Using 1 and 2 , we get:

1.731\ r^+=0.731\times \frac {6.28}{2}

<u>Size of the potassium ion = 1.33 Å</u>

4 0
3 years ago
If a gas has a volume of 1000 ML at a temperature of 23°C and a pressure of 100 mmhg, what is it’s volume under standard conditi
Colt1911 [192]

Answer:

119.7 mL.

Explanation:

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<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

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8 0
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kicyunya [14]

Answer:

b.

Explanation:

Cation and anion.

For example Na+ and Cl-  -----> NaCl.

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3 years ago
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