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▹ Answer
<em>x¹⁰z¹⁵</em>
▹ Step-by-Step Explanation
x¹²z¹¹/x²z⁴
<u>Reduce</u>
x¹⁰z¹¹ * z⁴
<u>Calculate</u>
x¹⁰z¹⁵
Hope this helps!
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The molarity of aqueous lithium bromide, LiBr solution is 0.2 M
We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.
- Volume = 10 mL = 10 / 1000 = 0.01 L
- Molarity of Pb(NO₃)₂ = 0.250 M
- Mole of Pb(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Pb(NO₃)₂ = 0.25 × 0.01
Mole of Pb(NO₃)₂ = 0.0025 mole
Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂
Pb(NO₃)₂ + 2LiBr —> PbBr₂ + 2LiNO₃
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.
Therefore,
0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr
Finally, we shall determine the molarity of the LiBr solution
- Mole = 0.005 mole
- Volume = 25 mL = 25 / 1000 = 0.025 L
- Molarity of LiBr =?
Molarity = mole / Volume
Molarity of LiBr = 0.005 / 0.025
Molarity of LiBr = 0.2 M
Learn more about molarity: brainly.com/question/10103895
Answer:
The mass of SO2 will be equal to the sum of the mass of S and O2.
Explanation:
This can be explained by the <em>Law of Conservation of Mass</em>. This law states that mass can neither be created nor destroyed. Knowing this, we can say that the reactants of a chemical reaction must be equal to the products.
In this case, the reactants Sulfur (S) and Oxygen (O2) must equal the mass of the product Sulfur Dioxide (SO2). Therefore, the statement <em>"The mass of SO2 will be equal to the sum of the mass of S and O2" </em>is correct.
Answer:
Si las condiciones para que el magma permanezca líquido no perduran, el magma se enfriará y solidificará en una roca ígnea. Una roca que se enfría en el interior de la Tierra se denomina intrusiva o plutónica y su enfriamiento será muy lento, produciendo una estructura cristalina de granos grueso.
Explanation:
Answer:
Chitin
Explanation:
A fly's main body is made of a thorax and an abdomen. These parts are covered by a hard exoskeleton made of a bony substance called chitin.