Mud is a suspension because it is a heterogeneous mixture that has many solid particles that have sedimented in various parts of the mixture, meaning they can be easily seen or removed. A solution is homogeneous like putting sugar in water or similar.
The part of the experiment that’s is not touched by the independent variable and is for comparison is called the :
Control Group
Answer:
2.12atm
Explanation:
Boyle's Law: P1V1 = P2V2
Manipulate to solve for unknown: P2 = P1V1/V2
Substitute values: P2=(1.2atm)(4.6L)/2.6L
P2 = 2.1230769atm
Round to 3 sig figs to get 2.12atm
<span>The </span>abundance of a chemical element<span> is a measure of the </span>occurrence<span> of the </span>element<span> relative to all other elements in a given environment. Abundance is measured in one of three ways: by the </span>mass-fraction<span> (the same as weight fraction); by the </span>mole-fraction<span> (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the </span>volume-fraction<span>. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and </span>ideal gas<span> mixtures. Most abundance values in this article are given as mass-fractions.
</span>
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>