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stellarik [79]
3 years ago
14

The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are

gray, and the electrons are black. These particles are not drawn to scale.
Which element below could be an isotope of this atom?
Need Help Asap! please and thank you.

A.sodium-10
B.beryllium-10
C.boron-9
D.carbon-9

Chemistry
1 answer:
lara [203]3 years ago
8 0
The correct option is B.
Isotopes refers to those chemical compounds which have the same number of protons and electrons but different number of neutrons, so they end up having different mass numbers.  The diagram given above is that of beryllium, which has atomic number 4 and it has 2 electrons in its outermost shell. It has four protons [same as the number of electrons] and 5 neutrons. Beryllium 10, which is its isotope has four electrons, four protons and 6 neutrons. To get the number of neutron, remove the number of electrons from the number given in the option, that is, 10 - 4 = 6.
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6 0
3 years ago
During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.
Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL

Putting values in above equation, we get:

1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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if the density of copper is 8.9g/cm^3 and the Volume of the sample is 23.4 mL what would be the mass? ​
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