1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GarryVolchara [31]
3 years ago
9

If told to determine whether or an unknown substance is either salt or sugar what do you do to determine which it is explain?

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
3 0
If you were allowed to taste it sugar will be more sweeter whereas salt wouldnt. But if you were to touch it then sugar is more tinnier than salt.
You might be interested in
Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?
Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

7 0
3 years ago
Read 2 more answers
Using the picture below, once air mass A reaches location B, the weather conditions at location B will most likely become
olga2289 [7]
I think the answer is a
3 0
3 years ago
A _______ reaction is when a compound containing carbon and hydrogen (and sometimes oxygen) combines with oxygen gas to produce
ch4aika [34]
This is categorized as a combustion reaction.
6 0
3 years ago
PLZ HELP PLZ PLZ ILL MARK AS BRAINLIESTT!!!!
LuckyWell [14K]

Q.1-

Given,

mass - 10grams

volume - 24 cm³

density = mass/volume

density = 10/24

density = 0.416 g/cm³

Q.2-

Given,

mass - 700grams

volume - 1100cm³

density = mass/volume

density = 700/1100

density = 0.6363 g/cm³

5 0
3 years ago
What was the weight percent of water in the hydrate before heating?
DedPeter [7]
Data:

weight of water before heating = 0.349

weight of hydrate before heateing = 2.107

Formula:

Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100

Solution:

Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %

Answer: 16.6%
3 0
3 years ago
Other questions:
  • In the lab, you mix two solutions (each originally at the same temperature) and the temperature of the resulting solution decrea
    12·1 answer
  • If we take 2.2 grams of CO2, 6.02 X 1021 atoms of nitrogen and 0.03 gram atoms of sulphur , then the molar ratio of C, N, and O
    15·2 answers
  • At 189k a sample of gas has a volume of 32.0cm. what does the gas occupy at 242k
    15·1 answer
  • Do you guys have viruses wit this website
    8·2 answers
  • Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider a mixture of six nitrogen molecules and six hydrogen molec
    5·1 answer
  • What factor can limit the buffering capacity of a molecule?
    15·1 answer
  • State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of
    13·1 answer
  • If an atom has twelve electrons, how many electrons are in the n=1 energy level? How many electrons are in the n=2 energy level?
    5·1 answer
  • HELP ME! If I get an F on my test I’m getting kicked out:(
    12·1 answer
  • Covalent compounds are generally not very hard. Justify the statement.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!