Answer:
0.41 atm
Explanation:
A reaction between liquid reactants takes place at 10.0 °C in a sealed, evacuated vessel with a measured volume of 5.0 L. Measurements show that the reaction produced 13. g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.
Step 1: Given data
Temperature (T): 10.0 °CVolume of the vessel (V): 5.0 LMass of sulfur hexafluoride gas (m): 13. gStep 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 10.0 °C + 273.15 = 283.2 K
Step 3: Calculate the moles (n) of SF₆
The molar mass of SF₆ is 146.06 g/mol.
13. g × 1 mol/146.06 g = 0.089 mol
Step 4: Calculate the pressure (P) of SF₆
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/V
P = 0.089 mol × (0.0821 atm.L/mol.K) × 283.2 K/5.0 L = 0.41 atm
LaKeisha is measuring the density of a solid piece of metal using the graduated cylinder method. She initially measures a volume of water in the cylinder to be 3.28 mL. After placing the metal into the graduated cylinder, the new volume was 8.72 mL. The mass of the metal was 42.26 g on a top loading balance.
Required:
What is the density of the metal calculated to the correct number of significant figures?
Answer: 7.77 g/ml
Explanation:
Volume of cylinder with only water = 3.28 mL
Volume of cylinder with water and metal = 8.72 mL
Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)
=8.72-3.28
=5.44 ml
Mass of metal = 42.26 g
Formula of Density = [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]
i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]
Hence, the density of metal = 7.77 g/ml
Two volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.
Answer:
Explanation:
Nitric oxide is the gas NO, it reacts with oxygen as shown below;
2NO(g) + O2(g) -----> 2NO2(g)
Now the gas formed is the gas NO2 which is known to be reddish brown in colour.
A diagrammatic representation of this reaction is shown in the image attached to this answer.
Image credit: Chemlibretext
what is the formula for H-H
Answer:
H-H equation is written as follows:
pH=pK + log
{HCO3-}(base)
{H2CO3}(acid)
A particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.
Answer:
Explanation:
From the information given;
Consider using Lande's Interval rule which can be expressed as:
[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]
here;
[tex]j+1[/tex] = highest level of j
and
[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]5(j+1) = 3(j+2)[/tex]
[tex]5j+5 = 3j+6[/tex]
[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]
recall that:
[tex]j = |S-L| \ \to \ |S+L |[/tex]
So;
[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &
[tex]S+L = \dfrac{5}{2} --- (1)[/tex]
Using the elimination method, we have:
[tex]2S = \dfrac{6}{2}[/tex]
[tex]S = \dfrac{3}{2}[/tex]
Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)
[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]
[tex]L = \dfrac{2}{2}[/tex]
[tex]L = 1[/tex]
A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)
Answer:
0.229 cm³.
Explanation:
The following data were obtained from the question:
Volume (in in³) = 0.014 in³
Volume (in cm³) =?
1 in = 2.54 cm
Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:
1 in = 2.54 cm
Therefore,
1 in³ = 2.54³ cm³
1 in³ = 16.387 cm³
Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:
1 in³ = 16.387 cm³
Therefore,
0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³
0.014 in³ = 0.229 cm³
Thus, 0.014 in³ is equivalent to 0.229 cm³.
Why is observational evidence important in an experiment?
Answer:
Observational evidence is essential for investigating the way disease affects populations, the patterns and distribution of risk within them, and the emergence of trends in health and disease over time.
Answer:
It tests a prediction It supports the results. It asks a testable question It predicts what will happen
Explanation:
definition of solubility
(science)
Answer:
th relative ability of a solute to devolve into a solvent
Molecule A undergoes isomerization to molecule B in acetone. Using curved arrows, showing key intermediates and any formal charges, propose a detailed mechanism for this isomerization. Provide a brief explanation why this isomerization occurs.
Answer:
hello your question is incomplete attached below is the complete question
answer :
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
Explanation:
Reason for the mechanism
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
attached below is the detailed mechanism
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.
is C5H10 ionic or covalent?
You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MSuppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.
Answer:
[N₂] = 1.1M
Explanation:
Based on the chemical reaction:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Equilibrium constant, K, is defined as:
K = 5.93 = [NO]² / [N₂] [O₂]
Where [] are equilibrium concentrations of each specie
As initial concentrations are:
N₂ = 0.40mol / 0.500L = 0.8M
NO = 1mol / 0.500L = 2M
The equilbrium concentrations are:
[NO] = 2M - 2X
[N₂] = 0.8M +X
[O₂] = X
Replacing:
5.93 = [2 - 2X]² / [0.8+X] [X]
5.93 = 4 - 8X + 4X² / 0.8X + X²
4.744X + 5.93X² = 4 - 8X + 4X²
1.93X² + 12.744X - 4 = 0
Solving for X:
X = -6.9M → False solution. There are no negative concentrations
X = 0.3M. Real solution.
[N₂] in equilibrium is:
[N₂] = 0.8M +0.3M
[N₂] = 1.1M
What produces the magnetic force of an electromagnet?
O magnetic fields passing through the device
O static charged particles on the wire
O movement of charged particles through the wire
O positive and negative charges repelling each other
Answer:
movement of charged particles through the wire .
Explanation:
When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
The formula for serial dilution can be used to obtain the molarity of solution A, B , C.
For solution AM1V1 = M2V2
M2 = 0.100 M × 10 mL/250-mL
M2 = 0.004 M
For solution BM1V1 = M2V2
M2 = 0.004 M × 25 mL/100-mL
M2 = 0.001 M
For solution CM1V1 = M2V2
M2 = 0.001 M × 20 mL/500-mL
M2 = 0.00004 M
Learn more about serial dilution: https://brainly.com/question/2167827
How can you model the cycling of matter in the Earth system?
Answer:
The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.
Explanation:
Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
What is Earth system?Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.
Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).
These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.
Therefore, Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
To learn more about Earth, refer to the link:
https://brainly.com/question/1204146
#SPJ6
1. Each substance written to the right of the arrow in a chemical equation is a
(1 point)
O catalyst
O reactant
O precipitate
O product
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles
Answer:
2.9moles of hydrogen gas
Explanation:
convert liters to dm³
since 1liter= 1dm³
thus, 65.0liters = 65.0dm³
number of moles = volume given/22.4dm³
= 65.0/22.4
=2.9moles
0
Which is not one of Earth's layers?
A А
crust
B)
inner core
mantle
D
ocean
The ocean is not a part of Earth's layers.
Answer:
Ocean
Explanation:
According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in millimolars), the reaction rate (in micromolars/min) is
Answer:
The answer is "A"
Explanation:
Please find the complete question in the attachment file.
[tex]\to R(s)= \frac{As}{K+s}[/tex]
when the s in the approach, that is infinity R(s) tends
[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]
16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.
d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt
f. the number of cobalt atoms in 4.23 mol of cobalt
g. the number of cobalt atoms in 4.23 g of cobalt
Answer:
d. 9.95 × 10⁻³ mol
e. 249 g
f. 2.55 × 10²⁴ atoms
g. 4.32 × 10²² atoms
Explanation:
d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol
e. the mass of 4.23 mol of cobalt
The molar mass of cobalt is 58.93 g/mol.
4.23 mol × 58.93 g/mol = 249 g
f. the number of cobalt atoms in 4.23 mol of cobalt
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms
g. the number of cobalt atoms in 4.23 g of cobalt
First, we will calculate the moles of cobalt using the molar mass of cobalt.
4.23 g × 1 mol/58.93 g = 0.0718 mol
Then, we will calculate the number of cobalt atoms using Avogadro's number.
0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms
balance the following equation by oxidation reduction method FeSO4
+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O
Answer:
[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Explanation:
Identify the elements with oxidation state changes:
Oxidation states of iron, [tex]\rm Fe[/tex]:
[tex]+2[/tex] in [tex]\rm FeSO_4[/tex] among the reactants.[tex]+3[/tex] in [tex]\rm Fe_2(SO_4)_3[/tex] among the products.Change to the oxidation state: [tex]+1[/tex] (oxidation) for each [tex]\rm Fe[/tex] atom.Oxidation state of manganese, [tex]\rm Mn[/tex]:
[tex]+7[/tex] in [tex]\rm KMnO_4[/tex] among the reactants.[tex]+2[/tex] in [tex]\rm MnSO_4[/tex] among the products.Change to the oxidation state: [tex](-5)[/tex] (reduction) for each [tex]\rm Mn[/tex] atom.The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:
[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].
(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)
Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:
[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Find the unknown coefficients using the conservation of atoms.
Reactants:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in two [tex]\rm K_2SO_4[/tex] formula units.Therefore, among the products:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in one [tex]\rm K_2SO_4[/tex] formula unit.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
[tex]5 \times 3 + 2 + 1 = 18[/tex] sulfur [tex]\rm S[/tex] atoms in five [tex]\rm Fe_2(SO_4)_3[/tex] formula units, two [tex]\rm K_2 SO_4[/tex] formula units, and one [tex]\rm MnSO_4[/tex] formula unit.Reactants:
There are already ten [tex]\rm S[/tex] atoms in that ten [tex]\rm Fe(SO_4)_2[/tex] formula units. The other [tex]18 - 10 = 8[/tex] formula units would correspond to eight [tex]\rm H_2SO_4[/tex] molecules among the reactants of this reaction.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
There are [tex]8 \times 2 = 16[/tex] hydrogen [tex]\rm H[/tex] atoms in that eight [tex]\rm H_2SO_4[/tex] molecules.Therefore, among the products:
There would be [tex]16 / 2 = 8[/tex] molecules of [tex]\rm H_2O[/tex], with two [tex]\rm H[/tex] atoms in each [tex]\rm H_2O\![/tex] molecule.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
An atom has 81 electrons, 84 neutrons, and 82 protons. What element is this atom?
Answer:
Lead
Explanation:
The subatomic particles within an atom can be used to know the atom or element given.
Of particular interest is the number of protons within the atom.
The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.
So; If we know the number of protons within an atom, we can know the element.
The number of protons given is 82, the element is therefore lead.
Answer:
The atomic number of polonium is 84. The atomic number lead is 82.
Explanation:
How many moles of water can be formed from 0.57 moles of hydrogen gas?
Answer:
0.57 water
Explanation:
To solve this problem, we need to write the reaction expression first.
The reactants are oxygen gas and hydrogen gas.
They react to give a product of water
2H₂ + O₂ → 2 H₂O
Given that;
Number of moles of hydrogen gas = 0.57moles
From the balanced reaction expression;
2 moles of hydrogen gas produces 2 moles of water
So;
0.57mole of hydrogen gas will also produce 0.57 water
A species of desert plant produces flowers that only bloom at night. How does this enhance the survival of the species?
A. This allows the plants to conserve water and not bloom during the heat of the day.
B. This species relies on nocturnal animals like moths for pollination and reproduction
C. This species relies on moonlight for photosynthesis.
D.This allows flowers to stay closed durng the day when herbivores are more likely to eat them.
Od
Answer:
A. This allows the plants to conserve water and not bloom during the heat of the day
Explanation:
Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.
Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.
To determine the concentration of citric acid, you will need to titrate this solution with 0.100 M NaOH. You are given a 1.00 M NaOH stock solution and will need to make enough 0.100 M NaOH to perform 3 titrations. For each titration, you will use 20.0 mL of 0.100 M NaOH solution.
Calculate the total volume (in mL) of the diluted solution you will need to prepare for the 3 titrations.
Determine the minimum volume (in mL) of 1.00 M NaOH stock solution needed to prepare the 0.100 M NaOH solution.
Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
PLZ HELP ASAP WILL GIVE BRAINLISTS TO RIGHT ANSWER
How many molecules of carbon dioxide are in 12.2 L of the gas at STP?
A) 3.28 x 10^23 molecules
B) 5.01 X 10^23 molecules
C)2.24 x 10^23 molecules
D)8.12 x 10^22 molecules
Answer:
c
Explanation:
ok than not c than b maybe
