Barium has a 2+ charge as it is in group 2 in the periodic table and so it has two electrons in its outer shell and chloride has a -1 charge on its chloride ion. So we will need two of the chloride ions as we have a 2+ charge to match the amount of charge on one barium ion- forming barium ion
BaCI2
Answer:
1.22 L of carbon dioxide gas
Explanation:
The reaction that takes place is:
- CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
First we <u>determine which reactant is limiting</u>:
- Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
- Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl
So HCl is the limiting reactant.
Now we calculate the moles of CO₂ produced:
- 0.05 mol HCl *
= 0.05 mol CO₂
Finally we use PV=nRT to <u>calculate the volume</u>:
- T = 25 °C ⇒ 25 + 273.16 = 298.16 K
1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
The objects which cannot be observed in detail without a microscope
include the following:
<h3>What is a Microscope?</h3>
A microscope is an instrument which is used to view smaller objects such as
microbe,cells, tissues etc. This instrument is used in viewing the different
cells found in the body as they can't be seen with the eye.
The remaining options which can be seen with the eyes don't require
the use of microscopes.
Read more about Microscope here brainly.com/question/25268499
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs
2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams
