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omeli [17]
2 years ago
14

How many neutrons would be in an atom of Selenium if the mass number for this particular isotope is Se-80?

Chemistry
1 answer:
EleoNora [17]2 years ago
6 0

Answer:2

Explanation:

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8. Fill in the blanks underneath the wave spectrum to indicate the relative positions of each type of
anzhelika [568]
Ultraviolet as ultraviolet light waves are around 420 nm like stated in the photo
7 0
3 years ago
All information on a line graph is as precise as the information in the data table. TrueFalse
Alexus [3.1K]
The answer to is all the information on a line graph is as precise as the information in the data table would be FALSE
4 0
3 years ago
What is the volume of an object with a density of 2.6 g/cm³ and a mass of 30.5g?
inessss [21]

Answer:

The answer is

<h2>11.73 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

mass of object = 30.5 g

Density = 2.6 g/cm³

The volume is

volume =  \frac{30.5}{2.6}  \\  = 11.7307692...

We have the final answer as

<h3>11.73 mL</h3>

Hope this helps you

5 0
2 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this
Salsk061 [2.6K]

The following are the steps  to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3                           N = +5                        

O = -2                            O = -2

Ag = +1                         Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

     NO2– -----> NO3–

     Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

     NO2– + H2O -----> NO3–

     Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

    NO2– + H2O -----> NO3– + 2H+

     Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     2 Ag+ + 2e------> 2 Ag(s)

3 0
2 years ago
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