C metallic bond I’m sure that that is the right answer
Answer:
0.33%
Explanation:
From the question given above, the following data were obtained:
Experimental value = 9.842 m/s²
Accepted value = 9.81 m/s²
Percentage error =?
The percentage error in the calculation can be obtained as follow:
Percentage error = |Experimental – Accepted| / Accepted value × 100
Percentage error = |9.842 – 9.81| / 9.81 × 100
Percentage error = 0.032 / 9.81 × 100
Percentage error = 0.33%
Therefore, the error in the calculation is 0.33%
The balanced chemical equation between calcium carbonate and hydrochloric acid:
![CaCO_{3}(s)+2HCl(aq)-->CaCl_{2}(aq)+H_{2}O(l)+CO_{2}(g)](https://tex.z-dn.net/?f=CaCO_%7B3%7D%28s%29%2B2HCl%28aq%29--%3ECaCl_%7B2%7D%28aq%29%2BH_%7B2%7DO%28l%29%2BCO_%7B2%7D%28g%29)
Moles of
=![2.56g*\frac{1mol}{100.09g}=0.0256mol CaCO_{3}](https://tex.z-dn.net/?f=2.56g%2A%5Cfrac%7B1mol%7D%7B100.09g%7D%3D0.0256mol%20CaCO_%7B3%7D)
Moles of HCl=![250mL*\frac{1L}{1000mL}*\frac{0.125mol}{L} =0.03125mol HCl](https://tex.z-dn.net/?f=250mL%2A%5Cfrac%7B1L%7D%7B1000mL%7D%2A%5Cfrac%7B0.125mol%7D%7BL%7D%20%3D0.03125mol%20HCl)
The mole ratio of ![\frac{CaCO_{3} }{HCl} =\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BCaCO_%7B3%7D%20%7D%7BHCl%7D%20%3D%5Cfrac%7B1%7D%7B2%7D)
But we have
per 0.03125 mol HCl
Hence, HCl is the limiting reactant. The amount of product formed would depend on moles of HCl.
Mass of
=0.03125mol HCl *![\frac{1mol CaCl_{2} }{2mol HCl} *\frac{110.98 g}{1 mol}=1.73 g CaCl_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1mol%20CaCl_%7B2%7D%20%7D%7B2mol%20HCl%7D%20%2A%5Cfrac%7B110.98%20g%7D%7B1%20mol%7D%3D1.73%20g%20CaCl_%7B2%7D)
There are 3 significant figures
The Ammonia present in 2.5 mole of Ammonia is 42.5grams.
<h3>What is a mole ?</h3>
A mole is a measuring unit in Chemistry to measure the number of atoms in certain molar mass of a substance.
It is given that 2.5 mole of Ammonia is present
Grams of Ammonia present in 2.5 mole of Ammonia = ?
Molecular weight of Ammonia = 17 grams.
1 mole of Ammonia = 17 grams of Ammonia
2.5 mole of Ammonia = 17 * 2.5
= 42.5 grams
Therefore 42.5 grams of Ammonia is present in 2.5 mole of Ammonia.
To know more about Mole
brainly.com/question/26416088
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