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aleksandr82 [10.1K]
2 years ago
14

Consider the nuclear equation below. Superscript 124 subscript 56 B a right arrow superscript 124 subscript 55 upper C s plus qu

estion mark. Which completes the nuclear equation? Superscript 0 subscript plus 1 e. Superscript o subscript negative 1 e. Superscript 1 subscript 0 n. Superscript 1 subscript 1 upper H..
Chemistry
2 answers:
nadezda [96]2 years ago
6 0

Superscript o subscript negative 1 e.

Explanation:

The nuclear reactions is of 2 types, one is nuclear fusion and the other one is nuclear fission.

Nuclear fusion is nothing but the combining of 2 nuclei with an emission of energy along with an electron, proton or beta particle.

Nuclear fission is the break down of a nucleus into 2 or more nuclei along with an electron, proton or beta particle.

And the reaction is,

₅₆B¹²⁴  ₅₅C¹²⁴ +  ₋₁e⁰

So the blank was filled by means of a beta particle.

VLD [36.1K]2 years ago
6 0

Answer:

The Answer is C.

Explanation:

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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

4 0
2 years ago
A pure gold bar is made of 19.55 mol of gold. What is the mass of the bar in grams
Law Incorporation [45]
<span>(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
19.55 x 196.97 = 
3850.76 g Au

I hope this helps you and have a great day!! :)
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4 0
2 years ago
Read 2 more answers
Nicolaas constructed a working model of the hydrologic cycle during science class. He used an open flame, liquid water, two larg
Nastasia [14]

Surface, condensation, evaporation, precipitation

3 0
2 years ago
Read 2 more answers
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

                                 P = 1.336 - 0.0094

                                P = 1.357 atm

7 0
2 years ago
Draw the Lewis structure for CH3CH2O2H?
kotegsom [21]

Answer:

hydroperoxyethane

Explanation:

tomsFor the Lewis structure we have to remember that all the atoms must have <u>8 electrons</u> (except for hydrogen). In this structure, we have three types of atoms, Carbon, Hydrogen and Oxygen. So, we have to remember the <u>valence electrons</u> for each atom:

-) Carbon : 4 electrons

-) Hydrogen: 1 electron

-) Oxygen: 6 electrons

We can start with the "CH_3" part. We can put 3 hydrogen bond arroun the carbon. We can use this same logic with  "CH_2". Finally for oxygens, we can put it one bond with CH_2 and a bond between oxygens with a final bond with hydrogen to obtain <u>hydroperoxyethane</u>.

See figure 1 for further explanations.

6 0
2 years ago
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