Question
2 Points
An object in a fluid will sink if:
A. the buoyant force is larger than the weight of the object.
B. the buoyant force is larger than the mass of the object.
C. the buoyant force is smaller than the weight of the object.
D. the buoyant force is smaller than the mass of the object.
SUBM

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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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:

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:

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.)

Similarly:

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.)

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:

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.)

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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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.

Answer:

[tex]\large \boxed{2 \, \%}[/tex]

Explanation:

1. Data

Mass of graduated cylinder              =  47.229 g

Mass of graduated cylinder + water =  71.821  g

Actual volume of water                     = 25.0     mL

2. Calculations

(a) Mass of water

Mass = 71.821 g -47.229 g  = 24.592 g

(b) Volume of water

[tex]\text{Volume} = \dfrac{\text{mass}}{\text{volume }} = \dfrac{\text{24.592 g}}{\text{ 1 g/mL}} = \text{24.592 mL}[/tex]

(c) Percent Difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{Measured - Calculated}\lvert}{ \text{Calculated}} \times 100 \,\%\\\\& = & \dfrac{\lvert 25.0 - 24.492\lvert}{24.492} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.5\lvert}{24.492} \times 100 \, \%\\ \\& = & 0.02 \times 100 \, \%\\& = & \mathbf{2 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{2 \, \%} }$}[/tex]

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 }

Answer:

C

Explanation:

A- incorrect, not in alphabetical order

B- incorrect, symbol for salt in NA not SA

C- correct

D- incorrect, not based on discovery

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

The periodic table, also known as the periodic table of the elements, is a tabular display of the chemical elements.

It is widely used in chemistry, physics, and other sciences, and is generally seen as an icon of chemistry.

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

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Answer:Crushed, decreased

Explanation:

Just got it right