Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
Best regards.
Iodine-131 is one of the most important isotopes used in the diagnosis of thyroid cancer. One atom has a mass of <u>130.906114</u> amu.\
<h3>What is
thyroid cancer?</h3>
Cancer that originates in the tissues of the thyroid gland is known as thyroid cancer. It is a condition where cells develop improperly and are susceptible to spreading to different bodily regions. A bump in the neck or swelling are examples of symptoms. Thyroid cancer is not always diagnosed because it can move from other parts of the body to the thyroid.
Young age radiation exposure, having an enlarged thyroid, and family history are risk factors. Papillary thyroid cancer, follicular thyroid cancer, medullary thyroid cancer, and anaplastic thyroid cancer are the four primary kinds. Ultrasound and tiny needle aspiration are frequently used in diagnosis. As of right now, it is not advised to screen those who are healthy and at normal risk for the disease.
To learn more about thyroid cancer from the given link:
brainly.com/question/11880360
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The earth has the moon captured in its gravity. this keeps the moon in an elliptical orbit
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V