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JulsSmile [24]
3 years ago
15

Diabetes occurs when there is a large amount of sugar in the blood. Which gland is most likely not functioning well enough in a

person who has diabetes?
pituitary
thyroid
pancreas
adrenal
Chemistry
1 answer:
Kisachek [45]3 years ago
6 0
The answer is the =Páncreas
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The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
Anyone want to talk ??​
3241004551 [841]
No is the answer your welcome well I don’t.
7 0
2 years ago
Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce
Iteru [2.4K]

Answer:

10.85 g of water

Explanation:

First we write the balanced chemical equation

3NO_{2} +H_{2}O -->2HNO_{3} +NO

Then we calculate the number of moles of nitric acid produced

n(HNO3) = \frac{mass}{molar mass} =\frac{75.9g}{63.02g/mol}=1.2044 mol

According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:

mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g

5 0
3 years ago
Please awnser the fill in blanks don’t mind the top or bottom pls help !!!!
zysi [14]
Can’t see the problems
7 0
3 years ago
What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?
SpyIntel [72]
When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) *                                           (1L ca(OH)2/0.585 mol Ca(OH)2 

                                          = 20.4 L
8 0
3 years ago
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