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Flura [38]
3 years ago
14

The isotope of iodine with mass number 131 is often used in medicine as a radioactive tracer. write the symbol for this isotope

indicating both mass number and atomic number.
Chemistry
1 answer:
vodka [1.7K]3 years ago
8 0
Iodine - 113, means iodine with mass number 113

To specify a particular isotope, add superscript and subscript on the left of symbol.

The superscript indicates the mass number (number of protons + neutrons)

The subscript indicates the atomic number (number of protons)

The iodine - 113 is writen

113
      I        [this the letter i]
 53

superscript 113 on the left ofd the symbol is the mass number = protons + neutrons

I is the symbol (letter i)

subscript 53 on the left of the symbol is the atomic number (number of protons)
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6:c

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Explanation:

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3 years ago
A simple circuit consists of a battery to provide power, wires to carry the ______, and load that uses the _________-for example
kipiarov [429]
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6 0
3 years ago
Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
3 years ago
Explain the reason potassium was visible when using the cobalt glass. Describe what occured.
Korolek [52]

Explanation:

If potassium is burnt the ions go into a high state of energy. Once they cool, it gives off energy in the form of a visible spectrum which has a characteristic color Now, The cobalt glass blocks out yellow light, and potassium ion which is purple in color is visible.

3 0
3 years ago
1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →
Leno4ka [110]

The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

  • The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

    If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

    But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.

  • 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

    As 80 g NaOH produces 106 g of Na₂CO₃.

    Therefore 5 g NaoH will produce ;

    106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

#SPJ1

5 0
2 years ago
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