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melamori03 [73]
2 years ago
9

Potassium impart purple colour but beryllium do not impart any colour to the flame.Why?​

Chemistry
1 answer:
BaLLatris [955]2 years ago
6 0

Answer:

<h2>The electrons in beryllium and magnesium are too strongly bound to get excited by flame. Hence, these elements do not impart any color.</h2>
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In the electron cloud model of the atom, an orbital is defined as the most probable(1) charge of an electron
Galina-37 [17]

The correct answer would be 3.) Location of an electron
3 0
3 years ago
Please answer this question ASAP
Nezavi [6.7K]

Answer:

put a test tube over the opening, remove it and quickly put a lit splint near the mout or in the tube. if you hear a squeaky pop it is hydrogen.

Explanation:

hydrogen ignites in air.

4 0
2 years ago
Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia
Wittaler [7]

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

           1.375       :           2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

4 0
3 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
3 years ago
Read 2 more answers
Need help with this.
Jobisdone [24]

Answer:

18.2 g.

Explanation:

You need to first figure out how many moles of nitrogen gas and hydrogen (gas) you have. To do this, use the molar masses of nitrogen gas and hydrogen (gas) on the periodic table. You get the following:

0.535 g. N2 and 1.984 g. H2

Then find out which reactant is the limiting one. In this case, it's N2. The amount of ammonia, then, that would be produced is 2 times the amount of moles of N2. This gives you 1.07 mol, approximately. Then multiply this by the molar mass of ammonia to find your answer of 18.2 g.

5 0
2 years ago
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