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2 years ago

How many moles are in 55.0 g of Aluminum? (Aluminum has a mass of 26.989)

Chemistry

1 answer:

Romashka [77]2 years ago

5 0

I think it’s C) 2.04

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During the process of sublimation,

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During this process. A. Solid turns directly into a Gas.

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3 years ago

Which element would mostly have chemical properties similar to that of magnesium (Mg)?

DanielleElmas [232]

The elements of same group are placed in same group as they have similar properties. So Calcium is the element which resembles Mangesium.

Calcium and Magnesium both are alkaline earth metals.

They have similar properties as they have similar outer or valence shell electronic configuration

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The electronic configuration of Mg [atomic number 12] and Ca [atomic number 20] are

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3 years ago

The chart below shows the major types of minerals mined in the United States and Australia.

Vlad1618 [11]

Fifty percent. plz brainliest my answer

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2 years ago

Read 2 more answers

Mr. Smith is hyperglycemic with a blood glucose level of 300mg/ml of blood. Explain how homeostasis would regulate his glucose l

aleksley [76]

Answer:

When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.

This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.

Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.

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2 years ago

A solution of acetic acid has a pH of 3.45. What is the concentration of acetic acid in this solution? Ka for CH3COOH is 1.8×10-

Serjik [45]

Answer:

0.007 M

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,

pH = - log [H⁺]

The expression of the pH of the calculation of weak acid is:-

$pH=-log(\sqrt{k_a\times C})$

Where, C is the concentration = ?

Given, pH = 3.45

So, for $CH_3COOH$ , $K_a=1.8\times 10^{-5}$

$3.45=-log(\sqrt{1.8\times 10^{-5}\times C})$

$\log _{10}\left(\sqrt{1.8\cdot \:10^{-5}C}\right)=-3.45$

C = 0.007 M

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3 years ago