Which image best represents the particles in liquids
Answer:
The 2nd Picture represents the particles in liquids.
Explanation:
Which elements cannot have more than an octet of electrons? Select all that apply
C
S
O
N
Br
Answer:
{ Carbon, Oxygen, Nitrogen }
Explanation:
Elements can only have more than an octet of electrons if they demonstrate an expanded octet. This is if they belong to groups in or beyond the third group. Why? Well these elements have d - orbitals that they can rely on to expand the number of electrons that could otherwise be limited. * Here we are focusing on main group elements, P - block elements more specifically. *
Carbon belongs to the 2 group, and thus doesn't have an empty d - orbital. Thus, it can't have more than an octet of electrons. Sulfur belongs to group 3, hence has an empty d - orbital, and can have more than an octet of electrons. Oxygen belongs to the 2 group, and thus doesn't have an empty d - orbital, so it can't have more than an octet of electrons. Same goes for Nitrogen. Bromine belongs to group 4, thus has empty d - orbitals, and can expand further than Sulfur can - it can have more than an octet of electrons.
Solution = { Carbon, Oxygen, Nitrogen }
Choose the species that is incorrectly matched with its electronic geometry.
1. BeBr2 : linear
2. CF4 : tetrahedral
3. NH3 : tetrahedral
4. H2O : tetrahedral
5. PF3 : trigonal bipyramidal
Answer:
PF3 : trigonal bipyramidal
Explanation:
PF3 has 4 domains around the central phosphorus (3 shared pairs and one lone pair of electrons), thus the electron geometry that has 4 domains is tetrahedral not trigonal bipyramidal
From the options the specie that is incorrectly matched is ( 5 ) ; PF₃ : trigonal bipyramidal
The specie PF₃ is composed of 3 shared pairs and one unshared pair of electrons ( i.e. It has 4 domains ) as seen in the Lewis structure of PF₃. therefore when writing its electronic geometry, it should expressed/written as tetrahedral and not trigonal bipyramidal.
Hence we can conclude that The specie that is incorrectly matched is PF3 : trigonal bipyramidal
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Carry out the following operations as if they were calculations of experimental results and express each answer in standard notation with the correct number of significant figures and wtih the correct units. Provide both the answer and the units.
1. 5.6792 m + .6 m + 4.33 m
2. 3.70 g - 2.9133 g
3. 4.51 cm x 3.6666 cm
Answer:
1. [tex]10.6\; \rm m[/tex] (one decimal place.)
2.[tex]0.79\; \rm g[/tex] (two decimal places.)
3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)
Explanation:
1.The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.
For example, in the first expression:
[tex]5.6792\;\rm m[/tex] has four decimal places.[tex]0.6\; \rm m[/tex] has only one decimal place.[tex]4.33\; \rm m[/tex] has two decimal places.Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)
Therefore:
[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)
2.Similarly:
[tex]\rm 3.70\; \rm g[/tex] has two decimal places.[tex]2.9133\; \rm g[/tex] has four decimal places.Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)
[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)
3.When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:
[tex]4.51\; \rm cm[/tex] has three significant figures.[tex]3.6666\; \rm cm[/tex] has five significant figures.Therefore, the result should have only three significant figures.
The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].
[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)
A buffered solution has a pH of 7.5. What would happen to the pH if a small
amount of acid were added?
Answer:
Dear user,
Answer to your query is provided below
When small amount of acid was added to buffered solution, pH will change very less.
Explanation:
Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.
A sample of carbon dioxide gas (CO2) contains 6 x 1022 molecules. How many moles of carbon dioxide does this represent?
Answer:
Explanation:
1 ST METHOD
GIVEN DATA:
number of molecules=6.022×10²²molecules
Avogodro's number Na=6.022×10²²
TO FIND:
number of moles
SOLUTION:
As we knom that number of moles=number of molecules/Na
number of moles=6.022×10²²/6.022×10²²
number of moles=1 moles
2ND METHOD:
AS WE KNOW THAT A MOLE OF A SUBSTANCE CONTAINS 6.022×10²² PARTICLES OF THAT SUBATANCE SO
1MOLE OF CO2=6.022×10²²MOLECULES OF CO2
what is the maximum number of electrons in p&q shell
Answer:
6 electrons
Explanation:
:)
hope this helps :))))))))
Answer:
each p shell can hold max. 6 electorns
i dont know what a q shell is
Write the chemical reaction for hydrogen thiocyanate in water, whose equilibrium constant is Ka. Include the physical states for each species
Write the chemical reaction for thiocyanate ion in water, whose equilibrium constant is Kb. Include the physical states for each species
Answer:
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
Explanation:
We identify the formula:
HSCN → hydrogen thiocyanate which is also known as Thiocyanic acid
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
As an acid, it gives proton to the solution. It is a weak acid, because the Ka
Ka = [SCN⁻] . [H₃O⁺] / [HSCN]
As a weak acid, the thiocyanate ion, will be the conjugate strong base. In water It can make hydrolisis:
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
As a base, it takes a proton from water.
Kb = [HSCN] . [OH⁻] / [SCN⁻]
H-S-C-N is the structure of hydro-thi-ocyanate.
H-S-C-N structure:1. combines with water, it produces the ion thiocyanate and the ion hydronium.
2. When thiocyanate combines with water, it produces the ion hydrogen thiocyanate and the ion hydroxy.
Find out more information about 'Thiocyanate'.
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Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if you eat 10,000 kJ (about 2390 kcal) one day and do nothing but sit relaxed for 14.3 h and sleep for the other 9.70 h
Answer:
[tex]m=107.8g[/tex]
Explanation:
Hello,
In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h of sleeping:
[tex]E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ[/tex]
Then, we compute the energy used that day:
[tex]E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ[/tex]
Finally, the mass by considering the consumed fat:
[tex]m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g[/tex]
Regards.
Express your answer to three significant figures.
This balanced equation shows the reaction of sodium hydroxide and sulfuric acid:
2NaOH + H2SO4 - Na2SO4 + 2H20.
In a laboratory experiment, a student mixes 355 grams of sulfuric acid with an excess of sodium hydroxide. What is the theoretical mass of
sodium sulfate produced? Refer to the periodic table and the polyatomic ion resource.
The theoretical mass of sodium sulfate is
grams.
Answer: The theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 355 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
First we have to calculate the moles of [tex]H_2SO_4[/tex].
[tex]\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{355g}{98g/mol}=3.62mol[/tex]
Now we have to calculate the moles of [tex]Na_2SO_4[/tex]
The balanced chemical equation is:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 1 mole of [tex]Na_2SO_4[/tex]
So, 3.62 mole of [tex]H_2SO_4[/tex] react to give 3.62 mole of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of [tex]Na_2SO_4[/tex]
[tex]\text{ Mass of }Na_2SO_4=\text{ Moles of }Na_2SO_4\times \text{ Molar mass of }Na_2SO_4[/tex]
Molar mass of [tex]Na_2SO_4[/tex] = 142 g/mole
[tex]\text{ Mass of }Na_2SO_4=(3.62moles)\times (142g/mole)=514g[/tex]
Therefore, the theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.
A sample of gas has a volume of 571 mL at a pressure of 4.04 atm. The gas is compressed and now has a pressure of 7.17 atm. Predict whether the new volume is greater or less than the initial volume, and calculate the new volume. Assume temperature is constant and no gas escaped from the container.
Answer:
The new volume is less than the initial volume.
The new volume is 322mL
Explanation:
Based on Boyle's law, the pressure of a gas is inversely proportional to its volume under constant temperature. That means if the pressure of a gas is increased, the volume decrease and vice versa. The formula is:
P₁V₁ = P₂V₂
Where P is pressure and V is volume of 1, initial state and 2, final states.
In the problem, the pressure of the gas increased from 4.04atm to 7.17atm, That means the new volume is less than initial volume because the gas is compressed occupying less volume.
Replacing in Boyle's equation:
4.04atm*571mL = 7.17atmV₂
322mL = V₂Beeing the new volume of the compressed gas 322mL
Temperature on Reaction Rate Use the drop-down menus to answer the questions. Which form of the sodium bicarbonate tablet has the most surface area? As the surface area increases, what happens to the average time required for the reaction?
Answer:Crushed, decreased
Explanation:
Just got it right
Why are covalent substances gases and liquid rather than solids?
Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond
hence covalent compounds exist as liquids, gases and soft solids
A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.582 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample?
Answer:
[tex]\% O=27.6\%[/tex]
Explanation:
Hello,
In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:
- Moles of carbon are contained in the 9.582 grams of carbon dioxide:
[tex]n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.218molC[/tex]
- Moles of hydrogen are contained in the 3.922 grams of water:
[tex]n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH[/tex]
- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:
[tex]m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO[/tex]
Finally, we compute the percent by mass of oxygen:
[tex]\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%[/tex]
Regards.
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg?
PLEASE HELP, will mark brainliest!!!
Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:
[tex]p_A=\chi_A\times P_T[/tex]
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = ?
[tex]\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}[/tex]
[tex]P_{T}[/tex] = total pressure of mixture = 525 mmHg
[tex]{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles[/tex]
[tex]{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles[/tex]
Total moles = 1.94 + 1.35 = 3.29 moles
[tex]\chi_{O_2}=\frac{1.94}{3.29}=0.59[/tex]
[tex]p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg[/tex]
Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:
[tex]mO_2 = 100 g - 37.8 g = 62.2 g[/tex]
We will convert both masses to moles using their molar masses.
[tex]N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol[/tex]
The mole fraction of O₂ is:
[tex]\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590[/tex]
Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.
[tex]pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg[/tex]
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
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The pressure of sulfur dioxide (SO2) is 2.27 x 104 Pa. There are 418 moles of this gas in a volume of 57.9 m3. Find the translational rms speed of the sulfur dioxide molecules.
Answer:
The translational rms speed of the sulfur dioxide molecules. vrms = 52.8 m/s
Explanation:
The root-mean-square speed is the measure of the speed of particles in a gas. It is given by the formula below;
vrms=√3RT/M
where vrms is the root-mean-square of the velocity, M is the molar mass of the gas in kilograms per mole, R is the molar gas constant, and T is the temperature in Kelvin.
Molar mass of SO₂ in Kg/ mole = (64/1000) Kg/mol = 0.064 kg/mol
R = 8.314 m³.Pa/K*mol
T = ?
Using PV = nRT to find T
T = PV/nR
Substituting for T in the rms formula
vrms = √(3R*PV/nR)/M
vrms = √3PV/nM
vrms = √3 * 2.27 * 10⁴ * 57.9)/418 * 0.064
vrms = 52.8 m/s
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units.
Answer:
25.0 grams is the mass of the steel bar.Explanation:
Heat gained by steel bar will be equal to heat lost by the water
[tex]Q_1=-Q_2[/tex]
Mass of steel= [tex]m_1[/tex]
Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]
Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]
Final temperature of the steel = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2= 105 g[/tex]
Specific heat capacity of water=[tex]c_2=4.18 J/g^oC[/tex]
Initial temperature of the water = [tex]T_3=22.00^oC[/tex]
Final temperature of water = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))[/tex]
On substituting all values:
[tex](m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}[/tex]
25.0 grams is the mass of the steel bar.Answer:
[tex]m_{steel}=24.9g[/tex]
Explanation:
Hello,
In this case, since the water is initially hot, the released heat by it is gained by the steel rod since it is initially cold which in energetic terms is illustrated by:
[tex]\Delta H_{water}=-\Delta H_{steel}[/tex]
That in terms of mass, specific heat and temperature change is:
[tex]m_{water}Cp_{water}(T_f-T_{water})=-m_{steel}Cp_{steel}(T_f-T_{steel})[/tex]
Thus, we simply solve for the mass of the steel rod:
[tex]m_{steel}=\frac{m_{water}Cp_{water}(T_f-T_{water})}{-Cp_{steel}(T_f-T_{steel})} \\\\m_{steel}=\frac{105mL*\frac{1g}{1mL}*4.18\frac{J}{g\°C}*(21.50-22.00)\°C}{-0.452\frac{J}{g\°C}*(21.50-2.00)\°C} \\\\m_{steel}=24.9g[/tex]
Best regards.