Answer:
it helped alot of dcientisd
Explanation:
Answer:
Next, Mendeleev began a text for inorganic chemistry (concerned with substances that are not organic, such as minerals), and the result, Principles__of Chemistry (two volumes, 1868–1870), would become the standard text for the field until early in the 20th century. His research for this book would also lead him to his most renowned work.
Explanation:
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5
please clear explain
Answer:
50000ppm and 0.855M.
Explanation:
ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters
A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.
To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:
mg NaCl:
5g * (1000mg / 1g) = 5000mg
L Solution:
100mL * (1L / 1000mL) = 0.100L
ppm:
5000mg / 0.100L = 50000ppm
To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:
5g * (1mol / 58.5g) = 0.0855moles NaCl
Molarity:
0.0855mol NaCl / 0.100L = 0.855M
Which two elements have similar characteristics?
Answer:
The two elements are FLUORINE AND CHLORINE. These two elements have similar characteristics because they belong to the same group in the periodic table. On the periodic table, elements with similar properties are grouped together in the same group. Both chlorine and fluorine belongs to the halogen group.
Explanation:
Hope it helps, some how.
Identify the types of information that are necessary to communicate with emergency responders. Select one or more: Any chemicals involved in an incident Any other hazards present in the lab The history of safety incidents in the lab How the incident happened
Answer:
How the incident happened
Any chemicals involved in an incident
Any other hazards present in the lab
Explanation:
Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.
The emergency responders are persons that helps in the case of emergency to deal with hazardous conditions. The information provided to the emergency responders helps in dealing with the condition.
The information that has to be provided to the emergency responder in case of a laboratory incident helps them to handle the situation. The questions about how the incident occurs, the presence of hazardous chemicals in the laboratory will help in avoiding the further hazard. The involvement of the chemical in the hazard will help to deal with the situation.
For more information about emergency responders, refer to the link:
https://brainly.com/question/20455351
Chromuate found inside which of the following structures of a eukaryotic cell?
Answer:
Chromuates are found in the chromosomes of the eukaryotic cells!!!!
Explanation:
I learned this in 7th grade!!!!
Mark Me Brainliest plzz
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F. b. Calculate the pH at the halfway points to equivalence, B and D. [Key
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. [tex]Na_{2} CO_{3}[/tex]
Major Species at B:
1. [tex]Na_{2} CO_{3}[/tex]
2. [tex]NaHCO_{3}[/tex]
Major Species at C:
1. [tex]NaHCO_{3}[/tex]
Major Species at D:
1. [tex]NaHCO_{3}[/tex]
2. [tex]H_{2}CO_{3}[/tex]
Major Species at E:
1. [tex]H_{2}CO_{3}[/tex]
Major Species at F:
1. [tex]H_{2}CO_{3}[/tex]
b) pH calculation:
At Halfway point B:
pH = pK[tex]a_{1}[/tex] + log[[tex]CO_{3}[/tex][tex].^{-2}[/tex]]/[H[tex]CO_{3}[/tex][tex].^{-1}[/tex]]
pH = pK[tex]a_{1}[/tex] = 6.35
Similarly, at halfway point D.
At point D,
pH = pK[tex]a_{2}[/tex] + log [H[tex]CO_{3}[/tex][tex].^{-1}[/tex]]/[H2[tex]CO_{3}[/tex]]
pH = pK[tex]a_{2}[/tex] = 10.33
the number that go before symbols and formulas in a chemicql equation are
A.SuperScripts
B.Subscripts
C.Catalysts
D.Coefficients
Conclusion
In this activity, thermal energy was transferred ( into / out of the hand warmer while it changed from
liquid to solid.
whats the question?
The Question A 1.80 x 10-2 kg block of metal has the following dimensions: 0.4981 inches by 0.531 inches, by 0.5839 inches. Determine the density of the block in g/mL. Using the provided list of metals and their respective densities, what is the metal's identity
Answer:
Zinc
Explanation:
m = Mass of block [tex]1.8\times 10^{-2}\ \text{kg}=18\ \text{g}[/tex]
Dimension of block is 0.4981 inches by 0.531 inches, by 0.5839 inches
[tex]1\ \text{inch}^3=16.3871\ \text{mL}[/tex]
Volume of block
[tex]V=0.4981\times 0.531\times 0.5839\times 16.3871\ \text{mL}[/tex]
Density is given by
[tex]\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{18}{0.4981\times 0.531\times 0.5839\times 16.3871}\\\Rightarrow \rho=7.112\ \text{g/mL}[/tex]
The density of the metal is [tex]7.112\ \text{g/mL}[/tex]. The metal here is zinc.
What mass of Na2SO4 will be formed by addition of 17.0 g of NaHCO3 in aqueous solution to
an aqueous solution containing 0.400 mol of H2SO4?
Mass of Na₂SO₄= 14.346 g
Further explanationGiven
17 g NaHCO₃
0.4 mol H₂SO₄
Required
mass of Na₂SO₄
Solution
Reaction
H₂SO₄ (aq) + 2 NaHCO₃ (aq) ⇒ Na₂SO₄(aq) + 2 CO₂(g) + 2 H₂O (l)
mol NaHCO₃ :
= mass : MW
= 17 g : 84 g/mol
= 0.202
Find limiting reactant = mol : coefficient
H₂SO₄ = 0.4 : 1 = 0.4
NaHCO₃ = 0.202 : 2 = 0.101
NaHCO₃ as a limiting reactant (smaller ratio)
Mol Na₂SO₄ based on mol NaHCO₃
From the equation, mol ratio NaHCO₃ : Na₂SO₄ = 2 : 1, so mol Na₂SO₄ :
= 1/2 x mol NaHCO₃
= 1/2 x 0.202
= 0.101
Mass of Na₂SO₄ :
= mol x MW
= 0.101 x 142,04 g/mol
= 14.346 g
how many moles of H2O are needed to react with a mole of 3.0 of Na?
Answer: 3.0 moles of water [tex](H_2O)[/tex] are needed to react with a mole of 3.0 of Na
Explanation:
The balanced chemical reaction of sodium with water is as follows:
[tex]2Na+2H_2O\rightarrow 2NaOH+H_2[/tex]
According to stoichiometry:
2 moles of sodium [tex](Na)[/tex] reacts with = 2 moles of water [tex](H_2O)[/tex]
Thus 3.0 moles of sodium [tex](Na)[/tex] reacts with = [tex]\frac{2}{2}\times 3.0=3.0[/tex] moles of water [tex](H_2O)[/tex]
3.0 moles of water [tex](H_2O)[/tex] are needed to react with a mole of 3.0 of Na
< Progress: 19/21 groups Due Jan 15 at 11:55 PM Use the References to access important values if needed for this question. The compound cobalt (II) sulfate forms a hydrate with seven water molecules per formula unit. What are the name and formula of the hydrate
Answer:
CoSO4.7H2O
Cobalt II tetraoxosulphate VI heptahydrate
Explanation:
According to IUPAC nomenclature, compounds are named systematically.
We were told that there are seven water molecules in each formula unit Hence the correct formula of the compound is CoSO4.7H2O
According to IUPAC system, the metal is first named, followed by its oxidation state in Roman numerals. Then the name of the anion is mentioned as well as the number of water molecules. We have to take into account the correct prefix signifying the number of molecules of water of crystallization.
Phosphorus tribromide decomposes to form phosphorus and bromine, like this:
4PBr3(g)-->P4(g)+6Br2(g)
Also, a chemist finds that a certain temperature at the equilibrium mixture of phosphorus tribromide, phosphorus, and bromine has the following composition:
Compound pressure at equiibrium
PBr3 97.4 atm
P4 99.2 atm
Br2 97.2 atm
Calculate the value of the equilibrium constant Kp for this reaction.
Answer:
Kp = 929551.4
Explanation:
First of all, we state the equilibrium:
4PBr₃ (g) ⇄ P₄(g) + 6Br₂(g)
In order to determine Kp, we need the partial pressure of each gas at equilibrium. Expression for Kp is:
{(Parial Pressure P₄) . (Partial Pressure Br₂)⁶} / (Partial Pressure PBr₃)⁴
Kp = 99.2 . 97.2⁶ / 97.4⁴
Kp = 929551.4
Take account that Kp can be also calculated from Kc.
Kp = Kc (RT)^Δn where Δn is the value for (final moles - initial moles) of any gas
5. Infer if a water molecule (H20) has two hydrogen atoms and one oxygen atom, how would you describe the make- up of a carbon dioxide molecule (CO2)?
(please help)
Answer:
The correct answer is - one carbon atom and two oxygen atoms.
Explanation:
Carbon dioxide is a gas that is present in the atmosphere of the earth. The chemical formula of carbon dioxide is CO₂. In this colorless gas, there is one molecule of carbon dioxide is made up of one atom of carbon covalently double bonded to the two atoms of the oxygens.
Thus, the make- up of a carbon dioxide molecule (CO2) includes one carbon atom and two oxygen atoms.
Can anybody answer this question of chemistry?
Answer:
Answer:A
Answer:AExplanation:
Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol
= 180g/molNumber of moles of Glucose = Mass/Molar Mass
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55
Answer:
Answer:A
Answer:AExplanation:
Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol
= 180g/molNumber of moles of Glucose = Mass/Molar Mass
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556
= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55
A 175 g piece of iron and a 175 g piece of aluminum are placed in a hot water bath so that they are warmed to 99.7 o C. The metal samples are removed and cooled to 21.5 o C. Which sample undergoes the greater heat change
Answer:
The Aluminium sample
Explanation:
From the question,
ΔQ of the iron = Cm(t₂-t₁)........................ Equation 1
Given: C = specific heat capacity of iron, m = mass of iron, t₁ and t₂ = initial and final temperature respectively.
Given: m = 175 g = 0.175 kg, t₂ = 21.5°C, t₁ = 99.7°C
Constant: C = 444 J/kgK.
Substitute into equation 1
ΔQ = 0.175(444)(21.5-99.7)
ΔQ = -6076.14 J
Similarly, for aluminium,
ΔQ' = c'm'(t₂-t₁)...................... Equation 2
Given: m' = 175 g = 0.175 kg,
Constant: 900 J/kgK
ΔQ' = 0.175(900)(21.5-99.7)
ΔQ' = -12316.5 J
Hence the aluminium sample undergoes the greater heat change
A chemist has a block of copper metal (density is 8.96 g/mL). They drop the metal into a graduated cylinder containing water, and find the volume change is 1.30 mL. What is the mass of the block, in grams
Answer:
11.648 g
Explanation:
From the question,
Note: an object immersed in a fluid, displaced an amount of fluid, equal to its volume.
Density = mass/volume.
D = M/V................................................................ Equation 1
Where D = density of the copper block, M = mass of the copper block, V = volume of the copper block.
make M the subject of the equation
M = D×V......................................... Equation 2
Given: D = 8.96 g/mL, V = 1.3 mL
Substitute into equation 2
M = 8.96×1.3
M = 11.648 g
Hence the mass of the block is 11.648 g
Insulin is a protein that is used by the body to regulate both carbohydrate and fat metabolism. A bottle contains 225 mL of insulin at a concentration of 20.0 mg/mL . What is the total mass of insulin in the bottle?
The total mass of insulin in the bottle : m = 4500 mg
Further explanationGiven
225 ml of insulin
The concentration 20 mg/ml
Required
The total mass of insulin
Solution
Density is the ratio of mass per unit volume
Can be formulated
ρ = m / V
m = ρ x V
Input the value :
m = 20 mg/ml x 225 ml
m = 4500 mg
Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
273'K and 760 mmHg atm. What would be the new volume? Hinto use the
following gas law equation (p1"V1) /T1 = (p2*V2)/72 "SHOW WORK *
28.15 ml
25.18 ml
21.85 mL
Particles of a liquid"
The new volume : 21.85 ml
Further explanationGiven
V1=25,0 ml
P1=725 mmHg
T1=298K is converted to
T2=273'K
P2=760 mmHg atm
Required
V2
Solution
Combined gas law :
[tex]\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}[/tex]
Input the value :
V2=(P1.V1.T2)/(P2.T1)
V2=(725 x 25 ml x 273)/(760 x 298)
V2=21.85 ml
For the reaction of hydrogen with iodine
H2(g) + I2(g) rightarrow 2HI(g)
relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide.
Answer:
[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]
Explanation:
Hello!
In this case, considering the given chemical reaction:
[tex]H_2(g) + I_2(g) \rightarrow 2HI(g)[/tex]
Thus, by applying the law of rate proportions, we can write:
[tex]\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}[/tex]
Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:
[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]
Best regards!
all organisms begins life as a ________ cell.
Answer:
All organisms begin life as a single cell
Explanation:
Single cell also did you know 15 minutes can save you 15% more on car insurance
Which of the following is an example of an electric force acting at a distance?
Answer:
The electric force acting at a distance can be observed when someone rubs a glass rod with something woolen then the rod is taken away from the woolen material. The piece of paper leaps into air and sticks on to the rod. This is due to the electrostatic force of attraction that the paper sticks on the glass rod.
Explanation:
Answer:
Pam rubs a glass rod with wool, then moves the rod over a pile of small paper scraps. The scraps leap into the air and stick to the rod.
1. How many moles of CO2 can be produced from a reaction of 10.0 moles CzHg?
2C₂H6 + 702 4 CO2 + 6H₂O
Answer:
[tex]20[/tex] moles of CO2 can be produced from a reaction of 10.0 moles C2H6
Explanation:
In this reaction -
2 moles of C₂H6 produces four molecules of Carbon dioxide (CO2)
So 1 mole of C₂H6 will produce [tex]\frac{4}{2} = 2[/tex] moles of Carbon dioxide (CO2)
Thus, 10 moles of C₂H6 will produce [tex]2 * 10 = 20[/tex] moles of Carbon dioxide (CO2)
HELP PLEASE
Instantaneous speed is the rate of motion at an instant in time. Based on this statement, a reasonable conclusion is that if and object's instantaneous speed does not chang over time then,________.
•the object is quickly speeding up
•the object is very gradually slowing down
•the object is moving at a constant speed
•the object has returned to its beginning position
Answer:
The object is moving at a constant speed.
Which of the following represents a compound made of five molecules
OCOS
OC205
Осо
5002
Sulphur Dioxide reacts with Oxygen gas to form Sulphur Trioxide. If 5.6 moles of SO2 reacts with excess Oz, how many moles of
SO3 will form?
SO2 + O2 -> SO3
4.1 The Chemical Equation
LEARNING OBJECTIVES
Define chemical equation.
Identify the parts of a chemical equation.
A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:
hydrogen reacts with oxygen to make water
We can represent this chemical change more succinctly as
hydrogen + oxygen → water
where the + sign means that the two substances interact chemically with each other and the → symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as
H2 + O2 → H2O
This is an example of a chemical equation, which is a concise way of representing a chemical reaction. The initial substances are called reactants, and the final substances are called products.
Unfortunately, it is also an incomplete chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.
What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you cannot change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you can do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.
To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:
H2 + O2 → 2H2O
The 2 in front of the formula for water is called a coefficient. Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem—simply go back to the reactant side of the equation and add a coefficient in front of the H2. The coefficient that works is 2:
2H2 + O2 → 2H2O
There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more “back and forth” work when balancing a chemical equation.
12. Which gas has the greatest kinetic energy at STP?
a. He
b. Ne
c. Ar
d. none of the above (all have the same kinetic energy)
Answer:
d. none of the above (all have the same kinetic energy)
Explanation:
The kinetic theory of gases states that the molecules of an ideal gas experience a constant random motion.
At standard temperature and pressure (STP), the kinetic energy of an ideal gas such as hydrogen, argon, neon, sodium, oxygen, helium, magnesium, beryllium, nitrogen, carbon, fluorine, chlorine etc are all the same.
The standard temperature and pressure (STP) of an ideal gas is 273K and 100 kPa.
Hence, all of the gases have the same kinetic energy at standard temperature and pressure (STP).
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Where, K.E represents kinetic energy measured in Joules.
M represents mass measured in kilograms.
V represents velocity measured in metres per seconds square.
In the equation
2 NaCl
+
Br2 -----> 2 NaBr +
Cl2
What are the large 2s called?
Answer:
coefficients
Explanation:
Decide whether the compound is ionic or molecular, if you can.
Compound is a soft waxy white solid that can be easily shaped and molded by hand. When the flame of an ordinary laboratory burner is held about away from the solid, it rapidly and extensively melts into a clear liquid.
Answer:
Molecular solid
Explanation:
Molecular solids have a very low melting point. This is because, they are composed of weak intermolecular forces hence the layers of the solid easily fall apart as the solid melts.
We can see here that the solid being considered melts easily. Hence we can conclude that it is actually a molecular solid.
What is the mass of 1.0 × 10^9 molecules of aspartame?
Answer:
294.3 g/mol b.
The mass of 1.0*10^9 molecules of aspartame is 2.943*10^11 g/mol
ASPARTAME
Aspartame is used as an artificial non-saccharide sweetener 200 times sweeter than sucrose, and is commonly used as a sugar substitute in foods and beverages.
CALCULATION
mass of one molecule of aspartame = 294.3g/mol
mass of 1.0*10^9 molecules of aspartame =294.3*(1.0*10^9) g/mol
=2.943*10^11 g/mol
refer https://brainly.com/question/25225559
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if you know an object's distance from the sun in kilometers,how can you find the distance in AU?
Answer:
We divide our value by 1.5×10⁸
Explanation:
AU is a unit to determine distance in the space.
We know that AU means astronomic unit.
The value for conversion is 1.5×10⁸ Km
1 AU = 1.5×10⁸ Km
Let's see the example:
One star is located from the sun, around 3.35×10¹⁰ Km away. To determine the distance in AU we use the conversion factor:
3.35×10¹⁰ Km . 1 AU / 1.5×10⁸ Km = 223.3 AU