Answer:
Tundra and desert ecosystem are different especially when talking about their climate. Tundra is situated in northern latitude and is very cold having a negative 34 degrees and has a permafrost while desert is often situated closer to equator and thus have hot climate yet they have a similar characteristic when it comes to vegetation or the plants as both experience a water scarcity thus plants are already adapted to live in water-barren environment.
Explanation:
sound waves? like what they do.
Answer:
A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.
HEY. HOPE THIS HELPS♡
A sample of gas with an initial volume of 9.35 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume of 2.84 L and warmed to a temperature of 310 K. What is the final pressure of the gas in atmospheres (atm)?
a. 4.97 atm.
b. 0.113 atm.
c. 5.95 atm.
d. 7.03x10^3 atm.
Answer:
D
Explanation:
how do sea breezes and land breezes affect local weather
The shielding of electrons gives rise to an effective nuclear charge, Zeff, which explains why boron is larger than oxygen. Estimate the approximate Zeff felt by a valence electron of boron and oxygen, respectively.
a. +5 and +8.
b. +3 and +6.
c. +5 and +6.
d. +3 and +8.
e. +1 and +4.
Answer:
b. +3 and +6.
Explanation:
Zeff = Z - S
The Z denotes the no of protons i.e. atomic number
S denotes the non-valence electrons
For boron,
the electronic configuration is 1s₂ 2s₂ 2p₄
Now
Z = 5, S = 2
So,
Zeff = 5-2
= +3
For O, the electronic configuration is 1s₂ 2s₂ 2p₄
So,
Z = 8, S = 2
= 8-2
= +6
Hence, the second option is correct
The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature.Time (s) [H2O2] (mol/L)0 1.00120 ± 1 0.91300 ± 1 0.78600 ± 1 0.591200 ± 1 0.371800 ± 1 0.222400 ± 1 0.133000 ± 1 0.0823600 ± 1 0.050 Assuming that the rate= -delta [H2O2]/delta t determine the rate law, integrated rate law, and the value of the rate constant. Calculate [H2O2] at 4000. s after the start of the reaction.
Answer:
Explanation:
From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]
This relates to the 1st order reaction rate, whereby:
The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]
here:
[A] = reactant concentration at time (t)
[A]_o = initial concentration for the reactant
k = rate constant
As such, the order of the reaction is the first order
Rate constant [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]
Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]
The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]
From the given table:
the initial concentration of [tex]H_2O_2[/tex] = 1.00 M
∴
We can determine the concentration of the reactant at 4000s by using the formula:
[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]
[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]
[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]
Finally, at 4000s: the average rate is:
[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]
Can someone please help with these 2?
Equilibrium shifts to the right.
OPTION A
Question 3 0 / 1 point How many grams of carbon atoms are present in a sample of C3H8 if there are 4.10 moles of hydrogen atoms in the sample
Answer:
18.45 g of C
Explanation:
This is a problem of rules of three:
1 mol of C₃H₈ contains 3 moles of C and 8 moles of H
If 8 moles of H are contained in 1 mol of propane
4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles
Now, If 1 mol of propane contains 3 moles of C
0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C
Let's convert the moles to mass:
1.5375 mol . 12 g /mol = 18.45 g
g 1.000 atm of oxygen gas, placed in a container having a pinhole opening in its side, leaks from the container 2.14 times faster than does 1.000 atm of an unknown gas placed in this same apparatus. Which of these species could be the unknown gas
The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
Question on the image
Answer: The mass of carbon dioxide required is 308 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of octane = 100.0 g
Molar mass of octane = 114.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of octane}=\frac{100.0g}{114.23g/mol}=0.875mol[/tex]
For the given chemical reaction:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
By the stoichiometry of the reaction:
2 moles of octane produces 16 moles of carbon dioxide
So, 0.875 moles of octane will produce = [tex]\frac{16}{2}\times 0.875=7mol[/tex] of carbon dioxide
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:
[tex]\text{Mass of carbon dioxide}=(7mol\times 44g/mol)=308g[/tex]
Hence, the mass of carbon dioxide required is 308 g
Can you look at the picture Look at the picture ASAP and help please?
Answer:
Volume of the reaction vessel is increased - shift to the left
The reaction is cooled down - shift to the right
H2 is added to the system - shift to the right
The pressure of the system is decreased - shift to the left
A catalyst is added to the system - no change
Water is removed from the system - shift to the right
Explanation:
When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.
When the volume of a reaction system is increased, the equilibrium position shifts in the direction in which there is the highest total volume. This is the left hand side.
Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.
Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.
Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.
A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.
Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.
Look at pictures and help please
Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
Explanation:
More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.
Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.
Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
In the picture this is my last question pls.
Answer:
Chromosomes and I think its too many
Explanation:
Why can light be treated like a particle?
Answer:
Light can be treated like particles because it is made of chunks like things called protons.
Please help! Thanks in advance!
Where is ur question ⁉️
Write the precipitation reaction for cobalt(II) hydroxide in aqueous solution: Be sure to specify the state of each reactant and product.
Answer:
The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:
CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)
Explanation:
Cobalt (ii) hydroxide is an inorganic compound consisting of cobalt (ii) ions, Co²+ and hydroxide ions, OH-. It is insoluble in water and the pure form known as the beta form is a pink-coloured solid. The impure form which incorporates other anions in its molecular structure is blue in colour and is ustable.
Cobalt (ii) hydroxide is formed as precipitate when an alkaline metallic hydroxide such as sodium hydroxide is mixed with an aqueous cobalt (ii) salt such as cobalt (ii) sulfate. The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:
CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)
Being a basic hydroxide, cobalt (ii) hydroxide neutralizes acids to form cobalt (ii) salts and water. For example: Co(OH)₂ (s) + H₂SO₄ (aq) ---> CoSO₄ (aq) + H₂O
Thus, cobalt (ii) hydroxide is soluble in acids.
Cobalt(II) hydroxide is used mostly as a drying agent for paints, varnishes, and inks. It is also useful in the preparation of other cobalt compounds.
What is the solute and solvent in a solution of salt water
Answer:
The solute is salt and the solvent is water
Explain:
Because salt is a component the dissolves in the solvent
when the sum 4.9965 + 2.11 + 3.887 is calculated, to how many decimal places should the answer be reported?
Answer:
the correct answer is four decimal places
Note: Please show all work and calculation setups to get full credit. T. he following may be used on this assignment: specific heat of (water=4.184 J/g oC; ice=2.03 J/g oC; steam=1.99 184 J/g oC); heat of fusion of water=80. cal/g; heat of vaporization=540 cal/g; 1cal=4.184J.
Calculate the energy required (in J) to convert 25 g of ice at -15 oC to water at 75 oC.
Answer:
16974J of energy are required
Explanation:
The energy required is:
* The energy to heat solid water from -15°C to 0°C using:
q = m*S*ΔT
* The energy to convert the solid water to liquid water:
q = dH*m
* The energy required to increase the temperature of liquid water from 0°C to 75°C
q = m*S*ΔT
The first energy is:
q = m*S*ΔT
m = Mass water = 25g
S is specific heat of ice = 2.03J/g°C
ΔT is change in temperature = 0°C - (-15°C) = 15°C
q = 25g*2.03J/g°C*15°C
q = 761.3J
The second energy is:
q = dH*m
m = Mass water = 25g
dH is heat of fusion of water = 80cal/g
q = 80cal/g*25g
q = 2000cal * (4.184J/1cal) = 8368J
The third energy is:
q = m*S*ΔT
m = Mass water = 25g
S is specific heat of water= 4.184J/g°C
ΔT is change in temperature = 75°C-0°C = 75°C
q = 25g*4.184J/g°C*75°C
q = 7845J
The energy is: 7845J + 8368J + 761J =
16974J of energy are required
A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula of mannitol is
Answer:
[tex]C_3H_7O_3[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:
[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]
Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:
[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]
Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:
[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]
To get:
[tex]CH_{2.34}O[/tex]
Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:
[tex]C_3H_7O_3[/tex]
Regards!
g a commercial product for treating injuries contains 35.0 g of MgSO4 in one bag and 250 mL of water in a seperate bag. When the bags are broken and their contents mixed, the temperature of the system changes. Calculate this temperature change
Answer:
he a real G
Explanation:
og tripple og
3. How many electrons are in the M shell?
Answer:
i believe the answer is D: 18
Explanation:
Answer:
18
M shell can actually hold 18 electron as you move higher atomic number
what is the molecular formula for this compound
Answer:
4
Explanation:
because it has 3 carbons and 6 hydrogen
hope this helps :)
What was one idea Dalton taught about atoms?
Explanation:
All atoms of one type were identical in mass and properties.
A sample of 10K gold contains the following: 10.0g gold, 4.0g silver, 5.0g copper, and 5.0g nickel. What is the percent gold in the sample?
Answer:
I don't no answer sorry
Explanation:
you follow me
HELPPP PLEASEEEEE
Name the following alkane molecule:
Answer:
5–bromo–9–chlorodecane
Explanation:
To name the compound given above, the following must be obtained:
1. The longest continuous carbon chain. This gives the parent name of the compound.
2. The substituent group attached to the compound.
3. Position of the substituent group.
4. Combine the above to obtain the name.
Now, we shall determine the name of the compound as follow:
1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.
2. The substituent groups attached to the compound are:
I. Bromine (Br) => Bromo
II. Chlorine (Cl) => Chloro
3. The position of the substituent groups are:
I. Br => carbon 5
II. Cl => carbon 9
NOTE: numbering is done alphabetically.
4. Therefore, the name of the compound is:
5–bromo–9–chlorodecane
Answer:
A.
Explanation:
I chose this answer and it was correct ♀️
How many significant digits should be used to report the answer to each of the following calculations? (2.75518 + 9.01 + 3.3349) / (2.1)
Answer:
2
Explanation:
You write your answer with the same number of significant figures as the number with the smallest amount of figures, in this case, that number is 2.1 so your answer should be written with 2 significant figures.
Homologous chromosomes pair up during prophase I to form a ________________.
Fill it in PLSSSSSSSSSSSSS
i will give brainlist
List the safety measures she should take to stay safe while she carries out her experiment
Answer:
Wearing gloves
Explanation:
it helps on not touching chemicals
Complete the following equations (note that the equations are not balanced). Use the act
necessary
03
Li > K> Ba > Sr> Ca > Na > Mg > Al > Mn > Zn > Cr> Fe > Cd >
Co > Ni > Sn > Pb > H > Sb> Bi > Cu > Ag > Pd > Hg > Pt > Au
K1+Pb(NO3)2 →?
O KNO3 + Pbl2
O KNO3 + Pbl
OKNI + PbO2
Answer:
Explanation:
a
Can someone help me out with this please
Answer:
[tex]molar \: mass \: of \: copper(ii)nitrate = 64 + (14 + 48) \times 3 \\ = 250 \: g \\ 64 \: g \: of \: copper \: produces \: 250 \: g \: of \: copper \: nitrate \\ 10.36 \: g \: of \: copper \: will \: produce \: ( \frac{10.36 \times 250}{64} ) \: g \\ = 40.7 \: g[/tex]