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Softa [21]
3 years ago
5

What is a common use of bases?

Chemistry
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

it’s D. to reduce indigestion

Explanation:

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In this lab, you will use the flame test to identify the metal ion in two boxes of unidentified fireworks. Write an investigativ
Rasek [7]
In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis. 

<span>Following are the investigative question that you can answer by doing this experiment.
</span>1) What information can be obtained from the colour of flame?
2) <span>State the relationship between wavelength, frequency, and energy?
</span><span>3) Can you identify the metal present in unknown sample provided?
4) How will you identify amount of metal present in sample solution?
5) </span><span>Why do different chemicals emit light of different colour?</span><span>

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6 0
3 years ago
Read 2 more answers
What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according
ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M  = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0
3 years ago
Helpppp<br> pleasee!!!!! :)
frez [133]
Weight cost toxicity
4 0
4 years ago
Read 2 more answers
Brainiest if correctly answered in 5m
erastovalidia [21]

I will list them from alkaline with the lowest boiling point and alkaline with the highest.

1. C2H6 
2. C9H20
3. C11H24
4. C16H34
5. C20H42
6. C32H66
7. C150H302

I have taken a quiz similar to this before and can assure you this is correct and is primarily because of the number of Carbons and Hydrogens within this. More Carbons and Hydrogens causes Boiling Points to increase because of stronger bonds.

8 0
3 years ago
Elements in the same column of the periodic table have what in common?
Kipish [7]

Answer:

Elements in same column of periodic table have same properties.

Explanation:

The elements in the same group have same number of valance electrons thus have similar properties.

Consider the elements of group two i.e alkaline earth metals. All have two valance electrons and show similar properties.

Magnesium, barium, calcium etc.

All alkaline earth metals form salt with halogens.e.g,

Mg   +   Cl₂    →    MgCl₂

Ba    +   Br₂    →     BaBr₂

Mg   +   Br₂    →     MgBr₂

Ca    +   Br₂    →     CaBr₂

They react with oxygen and form oxides of respective metal.

2Mg   +   O₂   →    2MgO

2Ba   +   O₂   →    2BaO

2Ca   +   O₂   →    2CaO

these oxides form hydroxide when react with water,

MgO  + H₂O   →  Mg(OH)₂

BaO  + H₂O   →  Ba(OH)₂

CaO  + H₂O   →  Ca(OH)₂

With nitrogen it produced nitride,

3Mg + N₂     →  Mg₃N₂

3Ba + N₂     →  Ba₃N₂

3Ca + N₂     →  Ca₃N₂

With acid like HCl,

Mg + 2HCl  →  MgCl₂ + H₂

Ba + 2HCl  →  BaCl₂ + H₂

Ca + 2HCl  →  CaCl₂ + H₂

8 0
4 years ago
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