Answer: Decreases the rate of reaction
- Remove water from food by dehydration.
- Transport food in a refrigerated truck.
- Store food in airtight containers.
- Store food in a refrigerator after opening.
Does not decrease the rate of reaction
- Store food in the open air.
- Place food on a warm surface.
Explanation: Dehydration of food excludes water from food which is one of the factor needed by microorganisms for growth, <em>so it decreaese the rate of reaction.</em>
Transporting food in refrigerated trucks lowers the temperature of food and not many microorganisms are active at very low temperatures, so it <em>decreases the rate of reaction.</em>
Storing food in airtight containers excludes air which is one of the factors required for microbial activity, so <em>it decreases reaction rate.</em>
Storing food in refrigerators after opening also <em>lowers the temperature of food and hence the the rate of microbial activit</em>y.
Storing food in the open air <em>does not decrease microbial activity</em> instead it provides microorganisms with the favorable conditions for their activity such as air and water from water vapor in the air.
Placing food on a warm surface <em>does not decrease rate of reaction</em> because microorganisms are very active in warm and humid environments.
Answer:
2
Explanation:
Mass of water molecule = mass of hydrated salt - mass of anhydrous salt
Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.
Number of moles = mass / molarmass
Molar mass of water = 18.015g/mol
No. of moles of water = 0.74 / 18.015 = 0.0411 moles.
Mass of BaCl2 present =?
1 mole of BaCl2 = 208.23 g
X mole of BaCl2 = 4.26 g
X = (4.26 * 1) / 208.23
X = 0.020
0.020 moles is present in 4.26g of BaCl2
Mole ratio between water and BaCl2 =
0.0411 / 0.020 = 2
Therefore 2 molecules of water is present the hydrated salt.
The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>
It matches the universal pH indicator and is indicating the proper pH
