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bearhunter [10]
2 years ago
8

Why is iodine placed just after tellurium on the periodic table even though its atomic mass is less than the atomic mass of tell

urium?
Chemistry
1 answer:
dem82 [27]2 years ago
3 0
Iodine follows Tellurium in the periodic table of the elements due to its higher atomic number. The atomic number (the number of protons in the nucleus) of Iodine is 53, while that of Tellurium is 52. The elements are not arranged by atomic mass.
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Describe how radioactive isotopes are used to treaat cancer​
stich3 [128]

Answer:

Radioisotope therapy is a procedure in which a liquid form of radiation is administered internally through infusion or injection. RIT's ultimate purpose is to treat cancerous cells with minimal damage to the normal surrounding tissue. These therapies are not normally the first approach used to fight a patient's cancer.

Explanation:

5 0
2 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
For full credit, show the details of each calculation below. Use more paper if needed.
vredina [299]

Answer:

Explanation:

1) a, b) A <em>solution</em><em> is a homogeneous mixture of two or more substances</em>. The <em>solute</em><em> is the substance present in a smaller amount</em>, and the <em>solvent</em><em> is the substance present in a larger amount. </em>

c) <em>A </em><em>saturated solution</em><em> contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. </em>

2) See picture in attachment.

8 0
2 years ago
The formation of a gas in a chemical reaction, due to ions changing places, often indicates that the reaction is which type?
Gelneren [198K]
A single-displacement reaction, also known as asingle-replacement reaction, is a type of chemicalreaction<span> where an element reacts with a compound and takes the place of another element in that compound. This type of </span>reaction<span> is typically pictured like this: Here, A replaces B in the compound BC.</span>
6 0
3 years ago
Read 2 more answers
If the reaction N2 (g) + 3 H2 (g) --&gt; 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo
Luba_88 [7]

<u>Answer:</u> The value of K_c is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The expression for the K_{c} is given as:

K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}

We are given:

[NH_3]=0.25M

[H_2]=0.75M

[N_2]=1.1M

Putting values in above equation, we get:

K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}

K_c=0.135

There are 3 conditions:

  • When K_{c}>1; the reaction is product favored.
  • When K_{c}; the reaction is reactant favored.
  • When K_{c}=1; the reaction is in equilibrium.

For the given reaction, the value of K_c is less than 1. Thus, the reaction is reactant favored.

Hence, the value of K_c is 0.136 and is reactant favored.

4 0
3 years ago
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