Answer:
Al is above H on the activity series
Explanation:
The given chemical reaction is expressed as:
2Al + 6HBr → 2AlBr₃ + 3H₂
The reaction above is a single displacement or replacement reaction and will only occur if Al is above H on the activity series.
The driving mechanism of such reaction is the position of the species on the activity series. The higher an atom is the more reactive and easier it will displace the lower ones.how you will separate pure water from a solution of water and salt
What is the empirical formula of a compound composed of 72.36% Fe and 27.64% O by mass?
empirical formula:
Answer:
Empirical formula = Fe₃O₄
Explanation:
Given data:
Percentage of Fe = 72.36%
Percentage of oxygen = 27.64%
Empirical formula = ?
Solution:
Number of gram atoms of O = 27.64 / 16 = 1.7
Number of gram atoms of Fe = 72.36 / 55.8 = 1.3
Atomic ratio:
Fe : O
1.3/1.3 : 1.7/1.3
1 : 1.3
Fe : O = 1 : 1.3
Empirical formula = 3( 1: 1.3)
Empirical formula = Fe₃O₄
what is not true about group 18
They are gases at room temperature.
They have 8 valence electrons.
They are known as halogens.
They are non-reactive and highly stable.
Answer:
they are known as halogens
Explanation:
everything else is correct
Which of these best explains why Earth's tectonic plates move?
b
Convection currents in the asthenosphere make fluid particles move.
Cooling causes contraction and shifting of the fluid particles in the lithosphere.
Cooling causes expansion and shifting of the fluid particles in the asthenosphere
d
Convection currents in the lithosphere make fluids move in a circular motion
Answer:
Cooling causes contraction and shifting of the fluid particles in the lithosphere.
Answer:
The 2 answer
Explanation:
bob hits a baseball.what two systems are working together to make this happen
Answer:
Muscular, and aerobic
Explanation:
the answer is bat and person
the most well-known biogenic d-elements
Answer:
Zn, Cu, Fe, Mn, Co, Mo
Explanation:
Biogenic elements are chemical elements constantly present in organisms and having definite biological significance(Bowen, 1966).
There are many of these biogenic elements in nature. Among the elements of the d-block, Zn, Cu, Fe, Mn, Co, Mo are worthy of mention among others.
Free ions of d- block metals are not found in the body. The elements are present in the form of bioinorganic complexes of the respective metals. They commonly exist as co-factors of proteins that carry out essential biological functions.
13. what is the precent composition
of Ca (C2 H3 O2) 2
Answer:Ca C2h3o2 2
Explanation
find the molar mass of the compound
ca : 1 X 40.1 = 40.1
c: 4 x 12.01 = 48.04
H: 6 X 1.01 = 6.06
O: 4 X 16 = 64
add these up and get 158.11
then divide 158.11 by each element massand multiply by 100
Ca: 40.01/158.11 X 100 = 25.3%
C: 30.4%
H:3.8%
O: 40.5%
How many grams of oxygen (O) are present in 0.0207 moles of Ca(HCO 3) 2
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
There are 285 mL in a cup of tea. What is the volume in liters?
Answer:
0.285 L
Explanation:
285ml divided by 1000=0.285L
The color of an outfit object is the color of the light that it
Reflects
Refracts I need help
Hello
Absorbs
it is still available please let us know
The rock has a mass of 6g and a volume of 3cm3. What is the density of the rock?
Answer:
2g/cm³
Explanation:
Given parameters:
Mass of rock = 6g
Volume of rock = 3cm³
Unknown:
Density of rock = ?
Solution:
Density is the mass per given volume of a substance. It can be mathematically expressed as:
Density = [tex]\frac{mass}{volume}[/tex]
Insert the parameters and solve;
Density = [tex]\frac{6}{3}[/tex] = 2g/cm³
what term is used to describe the formation of ions in an aqueous solution from a molecular compound
Answer:
Ionization
Explanation:
Molecular compounds are chemical compounds composed of discrete molecules. A molecular compound undergoes ionization when being dissolved in water and the formation of ions are being produced. For example, hydrogen chloride is a molecular compound, when it dissolves in water, ionization is being carried out, and ions are being formed.
[tex]\mathbf{HCl \to H^+_{(aq)} + Cl^-_{(aq)}}[/tex]
A voltaic cell is constructed that is based on the following reaction:
Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq).
If the concentration of Sn2+ in the cathode compartment is 1.00 M and the cell generates an emf of 0.16 V , what is the concentration of Pb2+ in the anode compartment?
Answer: The concentration of Pb2+ in the anode compartment is [tex]1.74\times 10^{-6}M[/tex]
Explanation:
[tex]Sn^{2+}(aq)+Pb(s)\rightarrow Sn(s)+Pb^{2+}(aq)[/tex]
Here Pb undergoes oxidation by loss of electrons, thus act as anode. Sn undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]
[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]
[tex]E^0=E^0_{[Sn^{2+}/Sn]}- E^0_{[Pb^{2+}/Pb]}[/tex]
[tex]E^0=(-0.14-(-0.13)V=-0.01V[/tex]
Now using Nernst Eqn :
[tex]E=E^0-\frac{0.059}{n}\log\frac{[Pb^{2+}]}{[Sn^{2+}]}[/tex]
[tex]0.16=(-0.01)-\frac{0.059}{2}\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]0.17=-0.0295\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]-5.76=\log\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex]1.74\times 10^{-6}=\frac{[Pb^{2+}]}{[1.00]}[/tex]
[tex][Pb^{2+}]=1.74\times 10^{-6}[/tex]
assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.0821 Lxatm/kxmol
Answer:
V = 22.41 L
Explanation:
Given data:
Mass of nitrogen = 14.0 g
Volume of gas at STP = ?
Gas constant = 0.0821 atm.L/mol.K
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles= 14 g/ 14 g/mol
Number of moles = 1 mol
Volume of gas:
PV = nRT
1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K
V = 22.41 atm.L / 1 atm
V = 22.41 L
Which among these were independent
variables?
Amount of sunlight
Amount of water
Amount of fertilizer
Plant growth
An unknown metal with a mass of 8.5 g was heated in boiling water to a temperature of 100°C. The metal was immediately transferred to an insulated cup containing 50.0 g of water at 22°C. At equilibrium (when the temperature became constant) the temperature of the system was 23.2°C. Calculate the specific heat of the metal and determine its identity. Explain how you arrived at your conclusion. You must show your work to receive credit for your answer.
The specific heat of the metal : 0.384 J/g° C,
and a metal with a specific heat of 0.384 is copper
Further explanationThe law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in = Q out
Q lost(metal) = Q gained(water)
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
[tex]\tt Q~metal=Q~water\\\\8.5\times c\times (100-23.2)=50\times 4.18\times(23.2-22)\\\\652.8\times c=250.8\Rightarrow c=\dfrac{250.8}{652.8}=0.384~J/g^oC[/tex]
How many different elements are involved in the chemical reaction shown?
2Fe2O2 + 3C --> 4Fe+3CO2
Answer:
3
Explanation:
fe is iron ,O is oxygen and C is carbon
How many moles of C are in a diamond that has a mass of 0.80 g? (Atomic mass of C=12.01g)
Answer:
0.066 moles
Explanation:
0.80/12.01=0.0666
1. At STP, potassium is classified as
A. a metallic solid
B.a nonmetallic solid
C.a metallic liquid
D. metalloid solid
List Earth’s four spheres.
Will give out free Brainiest
Answer:
The four spheres of it are,
Explanation:
1. Lithosphere
2. Atmosphere
3. Biosphere
4. Hydromasphere
May It's Help U
what is the value of 100cm^3 in dm^3
Answer:
0.1 dm³.
Explanation:
To obtain the value of 100 cm³ in dm³, do the following:
Recall:
1 cm³ = 0.001 dm³
Therefore,
100 cm³ = 100 cm³ × 0.001 dm³ / 1 cm³
100 cm³ = 0.1 dm³
Thus, 100 cm³ is equivalent to 0.1 dm³.
A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were transferred to 2.5 kg of water at 27˚C, what would the final temperature be?
The final temperature : 78.925°C
Further explanationHeat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Energy releases = 130 kcal = 130 x 4.18 kJ=543.4 kJ
The final temperature :
[tex]\tt \Delta T=\dfrac{Q}{m.c}\\\\\Delta T=\dfrac{543.4}{2.5\times 4.186~kJ/kg^oC}\\\\\Delta T=51.925^oC[/tex]
Final temperature :
ΔT=final-initial
51.925°c=final-27°c
final = 51.925+27=78.925°C
Which of the following elements should have properties similar to those of nitrogen
Answer:
Nitrogen group element, any of the chemical elements that constitute Group 15 (Va) of the periodic table. The group consists of nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), bismuth (Bi), and moscovium (Mc)
Explanation: Hope i helped! :D
What types of geological events can the following interactions produce?
Divergent boundaries:
Convergent boundaries:
Transform boundaries
Answer:
divergent boundaries- volcanic activity, shallow earthquakes, new sea floor.
convergent boundaries- earthquakes, volcanoes, formation of mountains.
transform boundaries- earthquakes, crustal deformation
Explanation:
(hope this helped!!)
he structure of the product, C, of the following sequence of reactions would be:
Answer:
I
Explanation:
The complete question can be seen in the image attached.
We need to understand what is actually going on here. In the first step that yields product A, the sodamide in liquid ammonia attacks the alkyne and abstracts the acidic hydrogen of the alkyne. The second step is a nucleophilic attack of the C6H5C≡C^- on the alkyl halide to yield product B (C6H5C≡C-CH3CH2).
Partial reduction of B using the Lindlar catalyst leads to syn addition of hydrogen to yield structure I as the product C.
Type of bond where two elements form oppositely charge ions that are attracted
Answer:
If electronegativity difference is greater than 1.7 then ionic bond other wise covalent bond
Explanation:
HELP ASAP I WILL GIVE BRAINLIEST
what is the duration of time in minutes for ammonium nitrate crystals and a pouch of water?
The temp range is 180*C to 250*C so how long can it last?
Answer:
it can last for 30 minutes
Explanation:
because it is very good at giving off heat, extothermal heat can last for quite a while.
In a chemical reaction, [_____] are the substances present after the reaction.
Answer:
Products
Explanation:
Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.80 g of methane is mixed with 1.92 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.
Answer:
1.08 g of water
Explanation:
The balanced chemical equation for the reaction of combustion of methane (CH₄) is the following:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)
According to the equation, 1 mol of CH₄ reacts with 2 moles of O₂. We convert from mol to grams by using the molar masses:
1 mol CH₄ = (1 x 12 g/mol) + (4 x 1 g/mol) = 16 g
2 mol O₂ = 2 x (2 x 16 g/mol) = 64 g
2 mol H₂O = 2 x ((2 x 1 g/mol) + 16 g/mol)= 36 g
From the masses of reactants (CH₄ and O₂), we can see that the stoichiometric ratio is 64 g O₂/16 g CH₄ = 4.
First, we have to identify which reactant is the limiting reactant. We can compare the stoichiometric ratio with the actual reactants ratio (the masses of reactants we have):
1.92 g O₂/0.80 g CH₄ = 2.4
As 4>2.4, we can conclude that O₂ is the limiting reactant.
Now, we consider the stoichiometric ratio between the limiting reactant (64 g O₂) and the product we have to calculate (36 g H₂O), and we multiply the ratio by the actual mass of O₂:
1.92 g O₂ x 36 g H₂O/64 g O₂ = 1.08 g
Therefore, 1.08 g of H₂O will be produced by the chemical reaction of 0.80 g of methane with 1.92 g of oxygen.
You are asked to write observations about a 100g sample of Nitrogen-16 before it decays , but you running late. In order to make observations, you require at least 5g of material Can make observations you are 57 seconds late to the lab?
Answer:
You can not make the observation in the lab
Explanation:
We must know the amount of Nitrogen-16 that remains after 57 seconds before we can make our conclusion.
Hence, from;
N/No = (1/2)^t/t1/2
Where;
N = amount of Nitrogen-16 present after 57 seconds
No = 100g of Nitrogen-16 originally present
t = 57 seconds
t1/2 = half life of Nitrogen-16 which is 7.13 seconds
Substituting values;
N/100 = (1/2)57/7.13
N/ 100 =(1/2)^8
N/100 = 1/256
256N = 100
N = 100/256
N = 0.39 g
You can not make the observation because you need at least 5g of Nitrogen-16 and only 0.39 g is left after 57 seconds.
The minimum amount of sample required has been 5 grams, while the remaining sample has been 0.39 grams. Thus, it has not been possible to make the observations.
The half life can be defined as the time required for the sample to reduce to half of its initial concentration. The half life can be given as:
Amount remained = Initial amount [tex]\rm \times\;\dfrac{1}{2}^\dfrac{time}{Half-life}[/tex]
The half life of nitrogen 16 has been 7.13 second.
The initial concentration of the sample has been 100 grams.
The concentration of sample remained after 57 seconds has been:
Amount remained = 100 [tex]\rm \times\;\dfrac{1}{2}^\dfrac{57\;sec}{7.13\;sec}[/tex]
Amount remained = 0.39 grams.
The amount of sample remained after 57 seconds has been 0.39 grams.
The amount of sample required for the analysis has been 5 grams, after 57 seconds only 0.39 grams of sample has been remaining. Thus, the student will not be able to made analysis of nitrogen-16.
For more information about the sample decay, refer to the link:
https://brainly.com/question/1831303