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Schach [20]
4 years ago
11

This is a group 2 element that is called the alkaline earth metals. It burns with a very bright light

Chemistry
1 answer:
erica [24]4 years ago
7 0

the alkaline eart metal s are

Beryllium

Magnesium

calcium

Strontium

barium


out of these Magnesium reacts with oxygen and produces bright light due to formation of MgO

all other metals also form oxides but the oxide of Mg is brightest

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Explain the difference between a suspension and a solution?
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Answer: A suspension is a heterogenous mixture containing large particles that will settle on standing. Sand in water is an example of a suspension. A solution is a homogenous mixture of two or more substances where one substance has dissolved the other.

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4 years ago
Convert 325 milliliters to liters.
steposvetlana [31]

Answer:

The Awnser Is A. 0.325 Liters

Explanation:

6 0
3 years ago
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A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the
lilavasa [31]

Answer:

<h2>B)</h2>

Explanation:

\text{Let}\\\\x-\text{a volume of}\ 20\%\ \text{saline solution}\\\\y-\text{a volume of }\ 50\%\ \text{saline solution}\\\\p\%=\dfrac{p}{100}\\\\20\%=\dfrac{20}{100}=\dfrac{2}{10}=0.2\\\\50\%=\dfrac{50}{100}=\dfrac{5}{10}=0.5\\\\0.2x-\text{a volume of salt in }\ 20\%\ \text{saline solution}\\0.5y-\text{a volume of salt in }\ 50\%\ \text{saline solution}\\\\60mL-\text{a volume of}\ 28\%\ \text{saline solution}\\\\28\%=\dfrac{28}{100}=0.28

0.28\cdot60mL=16.8mL-\text{a volume of salt in }\ 28\%\ \text{saline solution}\\\\\text{Equations}\\\\(1)\qquad x+y=60\\(2)\qquad0.2x+0.5y=16.8

\left\{\begin{array}{ccc}x+y=60\\0.2x+0.5y=168&\text{multiply both sides by 10}\end{array}\right\\\left\{\begin{array}{ccc}x+y=60&\text{multiply both sides by (-2)}\\2x+5y=168}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-2y=-120\\2x+5y=168}\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3y=48\qquad\text{divide both sides by 3}\\.\qquad y=16

\text{Put the value of}\ y\ \text{to the first equation:}\\\\x+16=60\qquad\text{subtract 16 from both sides}\\x=44

6 0
4 years ago
A 0.73 m solution of chloroacetic acid, , has a ph of 1.50. calculate for the acid.
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PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
6 0
3 years ago
A common cooking vinegar is 5.0% acetic acid (ch3cooh. how many molecules of acetic acid are present in 41.9 g vinegar?
UkoKoshka [18]
To calculate the number of molecules ch3cooh in a given amount of vinegar, we first need to determine the amount of ch3cooh in the solution by multiplying the percent <span>ch3cooh with the sample amount. Then, convert g to moles and use avogadro's number. We do as follows:

41.9 g vinegar solution ( .05 g </span><span>ch3cooh / g vinegar solution ) (1 mol / 60.05 g ) ( 6.022x10^23 g / 1 mol ) = 2.1x10^22 molecules </span><span>ch3cooh</span>
6 0
3 years ago
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