Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
Answer:
C. 81 c
Explanation:
took the test :)
What is the density of water (in g/mL) to fouf significant figures at 6°C?
Answer :
0.9999432 g/cm^3
The density of water in g/ml up to 4 significant figures at 6°C is 0.9999432 g/cm³.
What are significant figures?Significant figures are used for establishment of a number which is presented in the form of digits. These digits give a meaningful representation to the numbers.
The significant figures are the significant digits which convey the meaning according to the accuracy. These provide precision to the numbers and hence are called as significant numbers.There are rules for counting significant figures which are as follows:
1)All non-zero digits are significant .
2)All zeroes which occur between non-zero digits are significant.
3)All zeroes to the left and right of a non-zero digit are not significant.
4) All zeroes on right of decimal are significant if a non-zero number follows them.
5)All zeroes on right side of non-zero digit are significant
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Balancing oxidation-reduction reactions
Mg+ N2—>Mg3N2
Answer:
[tex]{ \sf{3Mg_{(s)} + N_{2(g)} →Mg _{3}N_{2(s)}}}[/tex]
Convert to standard notation.
9.3 x 10^-5
Answer:
9.3 x 10^ -5 in standard notation is 0.000093
Explanation:
Since 10 is raised to a -5, this means that there should be 5 zeros before the non-zero digits
Which is larger?
1). 1 liter or 1500 milliliters
2) 12 cm or 12 milliliters
Answer:
1). 1 litre
2).12 cm
Explanation:
In 1 litre = 1000 ml whereas 1500 milliliters equals 0.15 litres and it is same for 12 cm and 12 milliliters
GIVING 20 POINTS!! Apply Concepts Two chemists each plan an investigation in which they will combine hydrogen gas with chlorine gas to observe the chemical reactivity. The plans differ in that one will use 25 milligrams each of hydrogen and chlorine, and the other will use 1000 times this amount, i.e., 25 grams each of hydrogen and chlorine. Can the investigations reach similar conclusions about the reactivity of hydrogen with chlorine? Explain.
Answer:
Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2
Explanation:
1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios. This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction
2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.
3. A reaction with more reactants will need more care because more products are produced, which can be toxic
4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound
A. Identify the type of molecule shown in the drawing. (2 points)
Answer: 1) Caboxylic acid
2) Ether
3) Ester
4) Aldehyde
5) Amine
6) Alcohol
Explanation: The type of a molecule is based it's primary functional group.
https://en.wikipedia.org/wiki/Functional_group
The identities of the given molecules shown in the drawing are:
1) Carboxylic acid.
2) Ether
3) Ester
4) Ethane
5) Aldehyde
6) Amine
7) Alcohol
What are the organic compounds?Organic compounds are those compounds, which contain carbon-carbon and carbon-hydrogen bonds.
The five organic compounds are lipids, carbohydrates, protein, nucleotide, etc.
Thus, the correct option is 1) Carboxylic acid.
2) Ether
3) Ester
4) Ethane
5) Aldehyde
6) Amine
7) Alcohol
Learn more about organic compounds
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