Answer:
1. the group number of sodium is 1 and it is a metal
2. the group number of helium is 18 and it is a nonmetal
3. the group number of iodine is 17 and it is a nonmetal
4.the group number of calcium is 2 and it is a metal
5. lithium has similar properties to potassium
6. calcium has similar properties to magnesium
7. neon has similar properties to xenon
8. Iodine has similar properties to chlorine
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
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.
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.
< 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.
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.
Which of the following represents a compound made of five molecules
OCOS
OC205
Осо
5002
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
#SPJ2
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)
A 2.54 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder that contains 12.13 mL h2o. What will be the final volume reading on the graduated cylinder, in mL?
Answer: If I'm correct it's 12.54 mL but don't rely on this because I'm not sure
Explanation:
Answer:
12.49 ml
Explanation:
Density = mass / volume
Volume = mass / density
Volume of zinc = 2.54/7.14
Volume of zinc = 0.3557
Initial volume = 12.13 ml
Total volume = initial volume+ volume of zinc
Total volume = 12.13 + 0.3557
Total volume = 12.4857 = 12.49 ml
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
In the equation
2 NaCl
+
Br2 -----> 2 NaBr +
Cl2
What are the large 2s called?
Answer:
coefficients
Explanation:
What is the nickname for mitochondria
Answer:
the powerhouse of the cell
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
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
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
Write the numbers in ascending order: 7.88, 7.088, 7.808, 7.8, 8.7
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.
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.
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
What were the two main reasons that Dmitri Mendeleev's periodic table was more successful than Lothar Meyer's?
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:
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 student is making a glucose solution. The student is using 12.55 g of glucose (C6H1206) and a 500 mL volumetric flask. Use this information to calculate the molarity of the glucose solution.
Answer:
M = 0.138 M
Explanation:
Given data:
Mass of glucose = 12.55 g
Volume of solution = 500 mL
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 12.55 g/ 180.156 g/mol
Number of moles = 0.069 mol
Volume in L:
500 mL × 1 L /1000 mL
0.5 L
Molarity:
M = 0.069 mol / 0.5 L
M = 0.138 M
An irregularly-shaped piece of aluminum (Al) has a mass of 66.0 grams. What is the volume in cm³ of this piece of aluminum if its density is 2.70 g/cm³?
Answer:
[tex]V=24.45\ \text{cm}^3[/tex]
Explanation:
Given that,
Mass of a piece of Aluminium, m = 66 grams
The density of Aluminium, d = 2.70 g/cm³
We need to find the volume of this piece in cm³.
We know that,
Density = mass per unit volume
[tex]d=\dfrac{m}{V}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{66\ g}{2.7\ g/cm^3}\\\\V=24.45\ \text{cm}^3[/tex]
So, the volume of the piece is [tex]24.45\ \text{cm}^3[/tex].
A student measures 48.8 mL of concentrated nitric acid, which has a density of 1.55g/ mL. What is the mass of the acid?
The mass of the acid : 75.64 g
Further explanationGiven
volume = 48.8 ml
density = 1.55 g/ml
Required
the mass
Solution
Further explanation
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
Density formula:
[tex]\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}[/tex]
Input the value :
mass = ρ x V
mass = 1.55 g/ml x 48.8 ml
mass = 75.64 g
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.
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
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
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
When C2H4(g) reacts with H2O(g) to form CH3CH2OH(g) according to the following equation, energy is released into the surroundings. C2H4(g) H2O(g) CH3CH2OH(g) Is this reaction endothermic or exothermic
Answer:
Exothermic
General Formulas and Concepts:
Chemistry
Thermodynamics
Exothermic - releases heat as a productEndothermic - absorbs heat as a reactantExplanation:
We are given that the reaction takes place between C₂H₄ (g) and H₂O (g) to form CH₃CH₂OH (g) and is releasing heat.
This means that heat would be product as it is being released during the reaction.
Therefore, we have an exothermic reaction.
what is the molarity of a solutoin which contains 38.5 g of sodium chloride disolved in 325 ml solution ?
Answer:
Molarity = 2.02 M
Explanation:
Given data:
Molarity of solution = ?
Mass of sodium chloride = 38.5 g
Volume of solution = 325 mL (0.325 L)
Solution:
Number of moles = mass/molar mass
Number of moles = 38.5 g/ 58.44 g/mol
Number of moles = 0.658 mol
Molarity:
Molarity = number of moles / volume of solution in L
Molarity = 0.658 mol /0.325 L
Molarity = 2.02 M