Answer:
tissues,organs, organelles
how many moles are there in 2.20 x 10^23 molecules of Na2SO4
Answer:
Explanation:
450 grams of Na2SO4? 450 Na soullnol Na2SO4 16.02x1023 molec, Na₂SO4 = 119x1024. | 142.05g Na25041 Imol Naz
8. The density of a gas at 350 C is 0,087 g/L. Compute the density at STP.
Answer:
0.20 g/L
Explanation:
Step 1: Calculate the molar mass of the gas (M)
At T = 350 °C = 623 K and P = 1 atm (we will assume this data), the density (ρ) of the gas is 0.087 g/L. We can calculate the molar mass of the gas using the following expression.
ρ = P × M/R × T
M = ρ × R × T/P
M = 0.087 g/L × (0.0821 atm.L/mol.K) × 623 K/1 atm = 4.5 g/mol
Step 2: Calculate the density of the gas at STP
At standard temperature (T = 273.15 K) and standard pressure (P = 1 atm), the density of the gas is:
ρ = P × M/R × T
ρ = 1 atm × 4.5 g/mol /(0.0821 atm.L/mol.K) × 273.15 K = 0.20 g/L
Which of these is most likely made of pieces of rock that are weathered by water and wind?
O A diamonds
OB sand
OC grass
OD vegetables
Answer:
The answer would be sand.
Explanation:
Sand is weathered down tiny pieces of rock. Sand flows down rivers to shores and slowly builds up to form beaches.
Answer:
The Answer is gonna be B sand
Plz answer the first two questions and maybe the third. I will give brainliest.
Answer:
1. 24.45 moles
2. 437.90 grams
Bonus :
0.75 grams
Explanation:
Hope it was helpful ;)
Five identical test tubes are each filled from the following five copper (11) sulfate stock solutions. Which of the following test tubes would appear the lightest blue?
a) Stock solution made form 0.200 moles of CuSO4 dissolved to a total volume of 400 ml
b) Stock solution made form 0.150 moles of CuSO4 dissolved to a total volume of 300 mL
C)Stock solution made form 0.250 moles of CuSO4 dissolved to a total volume of 500 ml
d) Stock solution made form 0.175 moles of CuSO4 dissolved to a total volume of 400 ml
e) Stock solution made form 0.125 moles of CuSO4 dissolved to a total volume of 300 ml
Answer:
deez cutz
Explanation:
did i get it right
Calculate how many grams of sodium acetate you expected to make from your starting amount of sodium bicarbonate (0.5g). This is your theoretical yield.
Equation: NaHCO3 + HC2H3O2 = NaC2H3O2 + H2O + CO2
What is the energy of an electron in a Li+ ion when an electron moves from n = 2 to n =3?
Answer:
The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from
n
i
=
2
to
n
f
=
6
.
A good starting point here will be to calculate the energy of the photon emitted when the electron falls from
n
i
=
6
to
n
f
=
2
by using the Rydberg equation.
1
λ
=
R
⋅
(
1
n
2
f
−
1
n
2
i
)
Here
λ
si the wavelength of the emittted photon
R
is the Rydberg constant, equal to
1.097
⋅
10
7
m
−
1
Plug in your values to find
1
λ
=
1.097
⋅
10
7
.
m
−
1
⋅
(
1
2
2
−
1
6
2
)
1
λ
=
2.4378
⋅
10
6
.
m
−
1
This means that you have
λ
=
4.10
⋅
10
−
7
.
m
So, you know that when an electron falls from
n
i
=
6
to
n
f
=
2
, a photon of wavelength
410 nm
is emitted. This implies that in order for the electron to jump from
n
i
=
2
to
n
f
=
6
, it must absorb a photon of the same wavelength.
If aqueous solutions of Ba(OH)2 and HNO3 are mixed, what products are formed? Select one: a. BaN2(s) + H2O(l) b. Ba(NO3)2(aq) + H2O(l) c. Ba(s) + H2(g) + NO2(g) d. Ba2O(s) + NO2(g) + H2O(l) e. Ba3N2(s) + H2O(l)
Answer: Ba(NO3)2(aq) + H2O(l)
Explanation:
What is the pressure inside a container of 3 moles of gas with a volume of 60 Liters at a temperature of 400 K?
I just need the answer not a link please :)
Is anyone good at chemistry if so can someone help me please ?
(NO LINKS)
Question 15
We're given the [OH⁻] as 8.34 × 10⁻¹² M. Using the formula pOH = -log[OH⁻], the pOH of this solution would be -log(8.34 × 10⁻¹²) ≈ 11.08.
The pOH is, for lack of a better term, the "opposite" of pH: A pOH of 7 is neutral; a pOH less than 7 is basic; and a pOH greater than 7 is acidic.
This follows from the relation, pH + pOH = 14. In this case, with a pOH of 11.08, our pH would be 14 - 11.08 = 2.92, which is acidic (pH < 7).
Thus, the correct answer choice is B.
g Consider (12.5 A) micro-grams of a radioactive isotope with a mass number of (78 B) and a half-life of (32.6 C) million years. If energy released in each decay is 32.6 keV, determine the total energy released in joules (J) in 1 (one) year. Give your answer with three significant figures.
Answer:
Energy released = 18.985 J
Explanation:
The exponential decay of radioactive substance is given by -
N(t) = N₀ [tex]e^{-kt}[/tex]
where
N₀ = initial quantity
k = decay constant
Half life, [tex]t_{1/2} = \frac{ln 2}{k}[/tex]
⇒[tex]k = \frac{ln 2}{t_{1/2} }[/tex]
Given,
N₀ = 12.5 + 3 = 15.5 × 10⁻⁶ gm
[tex]t_{1/2}[/tex] = 32.6 + 18 = 50.6 × 10⁶ years
So,
[tex]k = \frac{ln 2}{50.6 * 10^{6} }[/tex] = 1.361 × 10⁻⁸ year⁻¹
Now,
N(1) = 15.5 × 10⁻⁶ [tex]e^{-1.361*10^{-8} *1}[/tex]
= 15.49999978904
Now,
Substance decayed = N₀ - N(t)
= 15.5 × 10⁻⁶ - 15.49999978904 × 10⁻⁶
= 21.095 × 10⁻¹⁷ kg
⇒Δm = 21.095 × 10⁻¹⁷ kg
So,
Energy released = Δmc²
= 21.095 × 10⁻¹⁷ × 3 ×10⁸ × 3 × 10⁸
= 189.855 ×10⁻¹
= 18.985 J
⇒Energy released = 18.985 J
What is the minimum temperature
needed to dissolve 35 grams of KCl in 100 grams of water?
Answer:
[tex]30^{\circ}\text{C}[/tex]
Explanation:
To know the temperature at which KCl dissolves in water we need to refer to the general solubility curves.
In the case of [tex]KCl[/tex], [tex]35\ \text{g}[/tex] of it will dissolve in [tex]100\ \text{g}[/tex] of water at a minimum temperature of [tex]30^{\circ}\text{C}[/tex].
So, the the minimum temperature needed to dissolve 35 grams of KCl in 100 grams of water is [tex]30^{\circ}\text{C}[/tex].
A compound with an approximate molar mass of
65.0g/mol is made up of C, H and Cl. This same
Compound contains 55% of Cl by mass .lf 9g
of the compound contains 4.19 x 10²³ atoms,
determine the compound's:
a empirical formular and molecular formular
Do not abuse or misuse any piece of drawing instrument. ASAPPPP
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62.5 mL of the base. The concentration of H2SO4 is ________ M. A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62.5 mL of the base. The concentration of H2SO4 is ________ M. 0.150 0.234 0.300 0.469 0.938
Answer: The concentration of [tex]H_2SO_4[/tex] is 0.234 M
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity [tex]H_2SO_4[/tex] = 2
[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = ?
[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 50.0 ml
[tex]n_2[/tex] = acidity of [tex]NaOH[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]NaOH[/tex] solution = 0.375 M
[tex]V_1[/tex] = volume of [tex]NaOH[/tex] solution = 62.5 ml
Putting in the values we get:
[tex]2\times M_1\times 50.0=1\times 0.375\times 62.5[/tex]
[tex]M_1=0.234M[/tex]
Therefore concentration of [tex]H_2SO_4[/tex] is 0.234 M
Check all that apply...helppppp
Answer:
dfgh
Explanation:
Help hurry please !!!!!
What is the reducing agent in the following reaction?
2 Br−(aq) + H2O2(aq) + 2 H+(aq) → Br2(aq) + 2 H2O(l)
Answer:
the reducing agent is Bromine
The oxidation state of an element is calculated by subtracting and the total sum of oxidation states of all the individual atom (excluding the one that has to be calculated) from total charge on the molecule. Bromine is the reducing agent in the following reaction.
What is oxidation state?Oxidation state of an element is a number that is assigned to an element in a molecule that represents the number of electron gained or lost during the formation of that molecule or compound.
The ionic equation is given as
2 Br⁻(aq) + H[tex]_2[/tex]O[tex]_2[/tex](aq) + 2 H⁺(aq) → Br[tex]_2[/tex](aq) + 2 H[tex]_2[/tex]O(l)
The oxidation state of bromine on reactant side is -1 while on product side it is 0 so, oxidation state of bromine has increased by 1 so, bromine is oxidized. If it is oxidized that means it must have reduced someone. So, bromine is acting as a reducing agent.
Therefore, bromine is the reducing agent in the given reaction.
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How is water used in society
Answer:
Water can be used for direct and indirect purposes. Direct purposes include bathing, drinking, and cooking, while examples of indirect purposes are the use of water in processing wood to make paper and in producing steel for automobiles. The bulk of the world's water use is for agriculture, industry, and electricity.
Explanation:
Answer:
water is used for drinking, bathing, cooking, agriculture etc
Which of the following astronomical bodies would most likely be the largest?
1. A dwarf star from a nearby solar system
2. A comet
3. One of the gas giants in our solar system
4. Ganymede, the largest moon of Jupiter
Answer:
1
because a dwarf star will seemlarge because of the in ability of any human being to see d sun
Make a
prediction about how a lack
of resources in an ecosystem
might impact the levels of
organization.
Answer:
Limiting factors of an ecosystem include disease, severe climate and weather changes, predator-prey relationships, commercial development, environmental pollution and more. An excess or depletion of any one of these limiting factors can degrade and even destroy a habitat.
Explanation:
The limiting factor of an ecosystem involve disease, weather change, climate change, environment pollution and more. An excess or depletion of any of these factor can destroy over habitat.
What is Ecosystem?An ecosystem is define as a community or a group of living organisms that live together and are dependent on each other.
There are two main types of ecosystem.
Terrestrial ecosystem-Terrestrial ecosystem are land based ecosystem and interaction of biotic and abiotic component in the specific area.
Example- forest, grassland desert etc.
Aquatic ecosystem-Aquatic ecosystem are ecosystem formed by surrounding water bodies. They are dependent on each other and their environment .
Example- lake, pond, river etc.
Thus ecosystem and their limiting factor have great impact on the level of organization.
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What is the maximum number of electrons that can be identified with each of following sets of quantum numbers? If there are none, enter 0.
Answer:
So, only one electron is there with n = 0, l = 0 and its quantum set is n = 0, l = 0, ml = 0, ms = -1/2. Hope, this is helping. The maximum number of electrons that can fit inside a 'n' shell is 2n^2. So it would be 1.
Explanation:
Quantum numbers are defined as the set of four numbers with the help of which we can get complete information about the electrons in an atom. Here the term 'n' represents the principal quantum number.
What is principal quantum number?The quantum number which represents the main energy level or shell in which electrons are present. It also determines the average distance of orbital or electron from the nucleus. It can have the whole number values like 1,2,3,4, ....
1. When n = 2, the maximum number of electrons present is 8. That is one 's' sub level and three 'p' sub levels. The spin of four electrons will be +1/2 and other four will be -1/2.
2. When l = 3, the possible orientations = 2l + 1 = 2(3) + 1 = 7. So the maximum number of electrons is 14.
3. The value of ml = -1 indicates only one orbital. So the maximum electrons is 2.
4. Here ml = -1, 0, 1 which shows three orbitals. So in ml = -1, there are only two electrons.
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What is the molar mass of AlCl3
Answer:
133.34 g/moles
Explanation:
but to make life easy it could be 133.4 g/ moles as well
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution. (b) Calculate the pH for the point at which 80.0 mL of the base has been added. (c) Calculate the pH for the equivalence point. (d) Calculate the pH for the point at which 105 mL of the base has been added.
Answer:
a. pH = 2.04
b. pH = 3.85
c. pH = 8.06
d. pH = 11.56
Explanation:
The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:
HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)
a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:
Ka = [H⁺] [NO₂⁻] / [HNO₂]
Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):
5.6x10⁻⁴ = X² / 0.15M
8.4x10⁻⁵ = X²
X = [H⁺] = 9.165x10⁻³M
As pH = -log [H⁺]
pH = 2.04b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:
pH = pKa + log [NaNO₂] / [HNO₂]
Where pH is the pH of the buffer,
pKa is -log Ka = 3.25
And [NaNO₂] [HNO₂] could be taken as the moles of each compound.
The initial moles of HNO₂ are:
0.100L * (0.15mol / L) = 0.015moles
The moles of base added are:
0.0800L * (0.15mol / L) = 0.012moles
The moles of base added = Moles of NaNO₂ produced = 0.012moles.
And the moles of HNO₂ that remains are:
0.015moles - 0.012moles = 0.003moles
Replacing in H-H equation:
pH = 3.25 + log [0.012moles] / [0.003moles]
pH = 3.85c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:
0.15M / 2 = 0.075M
The NaNO₂ is in equilibrium with water as follows:
NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺
The equilibrium constant, kb, is:
Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]
Where [OH⁻] = [HNO₂] = x
[NaNO₂] = 0.075M
1.79x10⁻¹¹ = [X] [X] / [0.075M]
1.34x10⁻¹² = X²
X = 1.16x10⁻⁶M = [OH⁻]
pOH = -log [OH-] = 5.94
pH = 14-pOH
pH = 8.06d. At this point, 5mL of NaOH are added in excess, the moles are:
5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH
In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =
3.66x10⁻³M = [OH⁻]
pOH = 2.44
pH = 14 - pOH
pH = 11.56How many mL of 0.100M Ca(OH) are needed to titrate 20.0mL of 0.300M H2SO4
Answer:
60.0 mL
Explanation:
Using CAVA = CBVB
CA = 0.300M
VA = 20.0 mL
CB = 0.100M
VB = ?
VB = CAVA
CB
VB = 0.300 * 20/ 0.100
VB = 60.0 mL
Worth 100 points plus ill mark brainliest
How many grams of sodium phosphate ( Na₃PO₄ )are required to make 125 milliliters of a 0.240 Molar solution?
4.92
6.48
8.44
12.5
Answer:
4.92 grams of sodium phosphate (Na₃PO₄) are required to make 125 milliliters of a 0.240 M.
Explanation:
Molarity is a measure of concentration that indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
[tex]Molarity=\frac{number of moles}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case:
Molarity= 0.240 Mnumber of moles= ?volume= 125 mL= 0.125 LReplacing in the definition of molarity:
[tex]0.240 M=\frac{number of moles}{0.125 L}[/tex]
Solving:
number of moles= 0.240 M*0.125 L
number of moles= 0.03 moles
Being the molar mass of sodium phosphate 164 g/mole, that is, the mass of one mole of the compound, you can calculate the mass of 0.03 moles using the following rule of three: if 1 mole of the compound has 164 grams, 0.03 moles contains how much mass?
[tex]mass=\frac{0.03 moles*164 grams}{1 mole}[/tex]
mass= 4.92 grams
4.92 grams of sodium phosphate (Na₃PO₄) are required to make 125 milliliters of a 0.240 M.
Calculate the enthalpy change for the photosynthesis of gluclose
Answer:
jhdgafhgafhagfhafg
Explanation:
Which do you use to qualify matter?
A. Thermometer
B. Five senses
C. Balance
D. Tape measure
Answer:
c balance
Explanation:
if you weigh something, you prove it's there.
3 attempts left
Check my work
Enter your answer in the provided box.
The pressure inside a 1.0 L balloon at 25°C was 750 mm Hg. What is the pressure (in mmHg) inside the
balloon when it is cooled to -65°C and expands to 3.3 L in volume?
mm Hg
Answer:
shhsss×<×>×××<××××
Explanation:
4×738×8<#329×
Put the steps of the carbon cycle in order using Step 1 as your starting point.
Step 1: Bacteria, through nitrogen fixation and nitrification, convert nitrogen into a usable form.
The animal dies and decomposes, returning nitrogen back to the soil.
Once nitrogen is in usable form, it is taken up by plants and assimilated into proteins..
An animal eats a plant and the nitrogen becomes part of the animal’s proteins.
Answer:
Carbon moves from the atmosphere to plants. ...
Carbon moves from plants to animals. ...
Carbon moves from plants and animals to soils. ...
Carbon moves from living things to the atmosphere. ...
Carbon moves from fossil fuels to the atmosphere when fuels are burned. ...
Carbon moves from the atmosphere to the oceans.
Explanation:
Answer:
step 4 , 2 , 3
Explanation:
