Answer: Thyroid disorders can range from a small, harmless goiter (enlarged gland) that needs no treatment to life-threatening cancer. The most common thyroid problems involve abnormal production of thyroid hormones. Too much thyroid hormone results in a condition known as hyperthyroidism.

Explanation:

4 0

3 years ago

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A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2KClO3 mc019-1.jpg 2KCI + 3O2 What is the perc

Anon25 [30]

The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%

3 0

3 years ago

Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of n-propanol (C3H7OH)is burned in air, -33.4 kJ of hea

Inga [223]

Answer : The heat of combustion of n-propanol is 0.554 kJ/mol

Explanation :

First we have to calculate the moles of n-propanol.

$\text{Moles of n-propanol}=\frac{\text{Mass of n-propanol}}{\text{Molar mass of n-propanol}}$

Molar mass of n-propanol = 60.09 g/mole

$\text{Moles of n-propanol}=\frac{1.00g}{60.09g/mole}=0.0166mole$

Now we have to calculate the heat of combustion of n-propanol.

As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ

So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)

= 0.554 kJ/mol

Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol

5 0

2 years ago