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velikii [3]
2 years ago
15

explain why some metal occur as free element in nature, why other occur in combined states. give two examples of each type

Chemistry
1 answer:
MaRussiya [10]2 years ago
3 0

Explanation:

Thus, those metals which remain unaffected by moisture, oxygen and carbon dioxide of the air can occur native or free. In other words, the unreactive metals occur in nature in free state because of their low reactivity towards chemical reagents. ... Metals usually occur in combination with nonmetallic elements.

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When ammonium carbonate decomposes what gas is produced and will a glowing splint burn brighter in the presence?
mestny [16]
When ammonium carbonate is heated it decomposes to give ammonia gas, carbon dioxide gas and water. The equation for the decomposition is;
(NH4)2CO3(s) = 2 NH3 (g) + CO2(g) + H2O(l)
A glowing splint would extinguish almost immediately because of the presence of carbon dioxide. Carbon dioxide does not support burning which is the property that makes it used a s a fighter extinguisher.
7 0
3 years ago
What are the correct coefficients when this chemical equation is<br> balanced? *<br> P4 + 02 P2O5
finlep [7]

Answer:

What are the correct coefficients when this chemical equation is

balanced? *

P4 + 02 P2O5

<h2>1, 5, 2</h2>

Explanation:

For this reaction we have a combination reaction. Balancing Strategies: This combination reaction is a lot easier to balance and if you can get an even number of oxygen atoms on the reactants side of the equation.

3 0
3 years ago
If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?
ipn [44]

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

5 0
2 years ago
MM H2O2 = 34.02 g/mol MM H2O = 18.02 g/mol MM O2 = 32 g/mol
Kaylis [27]

  The grams   of oxygen that  are   produced  is  228.8 grams


  <em>calculation</em>

2H₂O₂ → 2H₂O +O₂

Step 1:  use  the  mole ratio to determine the moles of O₂

from equation above H₂O₂:O₂  is   2:1

therefore the  moles of O₂  = 14.3 moles ×1/2 = 7.15   moles

Step 2:  find  mass   of O₂

mass = moles ×  molar mass

= 7.15 moles × 32 g/mol =228.8 g

3 0
3 years ago
What is the element name for K2S
aliya0001 [1]
~The picture below shall help :)
-Hope this helped ^_^

5 0
3 years ago
Read 2 more answers
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