Element A has 15 protons. Determine the number of neutrons in its isotope with mass number 33

Answer: 3 qualitative: the word pharmacology is written in blue, the stairs are gray, the tile on the first floor is yellow. 3 quantitative: there are four people wearing yellow, there are Four pink boxes, there are two floors.

Inference: there are many people performing experiments so it’s a lab.

Explanation:

Answer:

Qualitive - The bars are yellow, The floor is green, and the floor has pharmacology on it.

Quantitative - There are 8 support bars, There is 3 gray coats, and there is 27 pills that I can see on that picture.

Inference - There are many people doing experiments because the word pharmacology is on the floor.

Explanation:

This picture tells you that pharmacology is a form of study of medicine and drugs.

Answer:

A. 0.2395 w/w %

B. 2394ppm

Explanation:

A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:

Mass glycerol / Total mass * 100

Mass glycerol:

The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):

2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol

Mass of water:

998.9mL and density = 0.9982g/mL:

998.9mL * (0.9982g/mL) = 997.1g of water.

That means percent by mass is:

% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %

B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:

2.394g * (1000mg / 1g) = 2394mg:

Parts per million: 2394mg / L = 2394ppm

Considering the definition of percent by mass

A) the concentration of the glycerol solution in percent by mass is 0.2395%.

B) the concentration of the glycerol solution is 2394.34 ppm.

A) The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]mass percentage=\frac{mass of solute}{mass of solution} x100[/tex]

In this case, you have a 2.600×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.

So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:

number of moles of glycerol= 2.600×10⁻² M× 1 L

number of moles of glycerol= 2.600×10⁻² moles

Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:

2.600×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.39434 grams of glycerol

On the other side, the volume of water needed was 998.9 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:

998.9 mL×0.9982 [tex]\frac{g}{mL}[/tex]= 997.1 grams of water

Finally, the mass percentage of the solution can be calculated as:

[tex]mass percentage=\frac{mass of glycerol}{mass of glycerol + mass of water} x100[/tex]

Solving:

[tex]mass percentage=\frac{2.39434 grams}{2.39434 grams+ 997.1 grams} x100[/tex]

[tex]mass percentage=\frac{2.39434 grams}{999.49434 grams} x100[/tex]

mass percentaje= 0.2395 %

In summary, the concentration of the glycerol solution in percent by mass is 0.2395%.

B)  Parts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.

Being the mass of glycerol 2.39434 grams equal to 2394.34 mg (1 g=1000mg), the concentration is:

[tex]concentration=\frac{2394.34 mg}{1L}[/tex]

concentration= 2394.34 ppm

In summary, the concentration of the glycerol solution is 2394.34 ppm.

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Answer:

[tex]m_{HNO_3}=27.4gHNO_3[/tex]

Explanation:

Hello,

In this case, for the chemical reaction:

[tex]3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)[/tex]

The first step is to compute the theoretical yield of nitric acid via stoichiometry in terms of the 3:2 ratio between nitrogen dioxide (molar mass = 46 g/mol) and nitric acid (molar mass = 63 g/mol) respectively:

[tex]m_{HNO_3}^{theoretical}=40.0gNO_2*\frac{1molNO_2}{46gNO_2}* \frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3} \\\\m_{HNO_3}^{theoretical}=36.52gHNO_3[/tex]

Now, the actual amount is computed by taking into account the 75.0-% percent yield:

[tex]m_{HNO_3}=0.75*36.5gHNO_3\\\\m_{HNO_3}=27.4gHNO_3[/tex]

Best regards.

Answer: The correct answer would be : True

I hope that this helps you !

Answer: Pentane C5H12

Explanation:

The boiling point of a substance is simply defined as the temperature whereby a liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid will changes into vapor.

The likely molecular formula for this compound is Pentane i.e C5H12 due to the fact that its boiling point is between Butane with formula C4H10 and Hexane with formula C6H14 boiling points.

The molecular formula of the of the substance is [tex]\bold {C_5H_1_0}[/tex] (pentane).

Boiling point:

It is the temperature at which liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid start to into vaporize.

The boiling point of the given compound is 309 K which is between Butane  and Hexane.

Therefore, the molecular formula of the of the substance is [tex]\bold {C_5H_1_0}[/tex] (pentane).

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Answer:

octane

Explanation:

heptane: an alkane with 7 C's

nonane: an alkane with 9 C's

hexane: an alkane with 6C's

The name of the linear alkane that contains 8 carbons is octane. The correct option is C.

Saturated hydrocarbons are alkanes. This indicates that single bonds are used to connect each of their carbon atoms. They are not reactive. They react with oxygen, the process called combustion or burning. Examples are methane, ethane, etc.

Alkane is of many carbon chains. They can be single, double, triple, etc. The alkane that has 8 carbon atoms is called octane. A linear hydrocarbon is an octane. The bonds between each carbon in the molecular skeleton are limited to two.

5 carbon atoms are called pentane, six carbon atoms are called hexane, and seven carbon atoms are called heptane, as in series, the eighth carbon atom is called octane.

Thus, the correct option is C. octane.

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Answer:

Endothermic

Explanation:

Storing sugar for later use is an example of an endothermic reaction because that energy is being absorbed.

Answer:

There are 18572.9 arcminutes in 86% of a circle.

Explanation:

The degrees in 86% of a circle is:

[tex] degrees = \frac{86}{100}*360 = 309.6 ^{\circ} [/tex]

Now, we need to find the number of arcminutes in 309.6°:

[tex] arcminute = \frac{59.99 arcminute}{1 degree}*309.6 degrees = 18572.9 [/tex]

Therefore, there are 18572.9 arcminutes in 86% of a circle.

I hope it helps you!

Answer:

The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]

Explanation:

Two magnitudes are directly proportional when increasing one quantity increases the other in the same proportion or when decreasing one quantity decreases the other in the same proportion.

The rule of three allows the resolution of problems that are related to the proportionality of three known values ​​and a fourth value that is always an unknown. In other words, it is useful to establish the proportionality between 2 values ​​a and b through the knowledge of a third value c in order to calculate a fourth value x. In the case of direct margins, the rule of three between a, b and c and the unknown x is:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{c*b}{a}[/tex]

In this case, knowing that 1 meter is equal to 0.000621 miles, 1 second is equal to 0.000278 hours, the simple rule of three is applied as follows: if 1 meter is equal to 0.000621 miles, 2.998 * 10⁸ meters are equal to how many miles?

[tex]miles=\frac{2.998*10^{8} meters*0.000621 miles}{1 meter}[/tex]

miles=186,175.8

[tex]2.998*10^{8} \frac{m}{s} =\frac{2.998*10^{8}m}{s}=\frac{186,175.8miles}{s}[/tex]

Replacing the seconds by their equivalent in meters:

[tex]\frac{186,175.8miles}{s} =\frac{186,175.8miles}{0.000278 hours}= 669,697,122.3 \frac{miles}{h}[/tex]

Then:

2.998*10⁸ [tex]\frac{m}{s}[/tex] = 669,697,122.3 [tex]\frac{miles}{h}[/tex] ≅ 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]

The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]

Answer:

14.83mL

Explanation:

11.6/0.81=14.3

Answer:

...................................................................................

The mileage of the Toyota Prius hybrid in miles per gallon would be 49.39 miles/gallon.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.

As given in the problem a Toyota Prius hybrid gets 21 kilometers per liter in highway driving.

1 kilometers =  0.621 miles

21 kilometers = 21 ×  0.621 miles

                      =13.04 miles

1 liter =  0.264 gallon

The mileage in miles per gallon = 13.04 miles / 0.264 gallon

                                                     = 49.39 miles/gallon

Thus, the mileage of the Toyota Prius hybrid in miles per gallon would be 49.39 miles/gallon.

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Given :

Molarity of [tex]Na_3PO_4[/tex] , M = 0.213 M .

To Find :

How many ml of 0.213 M of [tex]Na_3PO_4[/tex] are required to deliver 66.4 mmol [tex]Na^+[/tex].

Solution :

Volume required  :

[tex]V=[\text{ 66.4 mmol of }Na^+] + [\dfrac{\text{1 mm of }Na_3PO_4}{\text{3 mmol of }Na^+}]+[\dfrac{\text{1 ml }Na_3PO_4}{\text{0.213 mmol of }Na_3PO_4}][/tex]

So ,

[tex]V=\dfrac{66.4}{3\times 0.213}\ ml\\\\V=103.91\ ml[/tex]

Therefore , volume of [tex]Na_3PO_4[/tex] required is 103.91 ml .

Hence , this is the required solution .

WOW! I have the same question! Again lol.

A similarity is that both are at the same temperature and a difference is both are at different states of matter between the particles in liquid water at 100ºC and the particles in steam at 100ºC.

There are only 3 states in which matter is present in the universe that is solid, liquid and gas, and on the basis of molecular space, they are differentiated in their respective category.

The water present both at 100ºC first is in a liquid state where molecules of water are more close as compared to the molecules of the steam water but the temperature for both is the same.

Therefore, the similarity is that both are at the same temperature and a difference is both are at different states of matter between the particles in liquid water at 100ºC and the particles in steam at 100ºC.

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Answer:

The answer is "[tex]\bold{0.525\ \ atm^{-1}}[/tex]"

Explanation:

Given equation:

[tex]2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)[/tex]

Given value:

[tex]\Delta rH =-198.2 \ \ \frac{KJ}{mol}[/tex]

[tex]Kp=1100 \ K[/tex]

[tex]\Delta x = 2-(2+1)\\\\[/tex]

     [tex]= 2-(2+1)\\\\= 2-(3)\\\\= -1[/tex]

[tex]\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right[/tex]

calculating the total pressure on equilibrium=  [tex](1-2x)+(0.5-x)+2x \ atm\\\\[/tex]

                                                                         [tex]= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\[/tex]

[tex]\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\[/tex]

calculating the pressure in  [tex]So_2[/tex]:

[tex]= (1-2 \times 0.15)[/tex]

[tex]= 1-0.30 \\\\ =0.70 \ atm[/tex]

calculating the pressure in  [tex]O_2[/tex]:

[tex]= (0.5- 0.15)\\\\= 0.35 \ atm \\[/tex]

calculating the pressure in  [tex]So_3[/tex]:

[tex]= (2 \times 0.15)\\\\= (.30) \ atm \\\\[/tex]

Calculating the Kp at 1100 K:

[tex]= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\[/tex]

Answer:

dispersion.

Explanation:

The molecule, PF2Cl3 is trigonal bipyramidal. The dipoles in the molecule cancel out since there is a symmetric charge distribution around the molecule hence the resultant dipole moment of the molecule is zero.

If the molecule is nonpolar, then the dominant intermolecular forces present are the weak dispersion forces, hence the answer above.

Answer:

[tex]V=23.5L[/tex]

Explanation:

Hello,

In this case, it is convenient to compute the car's mileage in km/L as follows:

[tex]25.0\frac{mi}{gal}*\frac{1km}{0.6214mi}*\frac{1gal}{3.78L}=10.64\frac{km}{L}[/tex]

In such a way, since the distance is measured to be 250 km, the volume requirement is:

[tex]V=\frac{250km}{10.64kg/L}\\ \\V=23.5L[/tex]

Regards.

Answer:

1. Ka = 8.16x10⁻⁵

2. pH = 2.40

Explanation:

1. The dissociation of a weak acid in water occurs as follows:

HA ⇄ H⁺ + A⁻

Ka = [H⁺] [A⁻] / [HA]

As 2% of the 0.20M solution is dissociated:

[H⁺] = [A⁻] = 0.20M * 2% = 0.004M -As H⁺ and A⁻ comes from the same reaction, their concentrations are the same

[HA] = 0.20M * 98% = 0.196M

Ka = (0.004)² / (0.196M) = 8.16x10⁻⁵

2. pH = -log [H⁺] = -log [0.004M]

want to talk

Explanation:

Explanation:

pure substances are those substances which are made up of one elements

mixture that those substances which are made by more than two or three elements

Answer:

Brainliest is appreciated

Explanation:

Substances which have a specific composition and cannot be separated into any constituents are called pure substances. Pure substances are further divided into elements and compounds. The combination of two or more pure substances is called a mixture. Mixtures can be classified into two types viz. heterogeneous and homogeneous mixtures. 

EXAMPLES

pure substance

gold, pure water, hydrogen gas

mixture substance

oil and water, sand and sugar

Answer:

B) The pH of a buffer solution does not change significantly when a small amount of acid is added.

C) The pH of a buffer solution is determined by the ratio of the concentration of acid to the concentration of base.

Explanation:

Buffer solution is useful in maintaining the pH of a solution constant . It is made of a weak acid and its salt or a weak base and  its salt . Ph of a buffer is given by the following relation .

pH = pKa + log [ A⁻ ] / [ HA ]

A- is a base and HA is a weak acid .

When we add acid , it reacts with A⁻ or base and gets neutralised .

When we add base , it reacts with acid HA and gets neutralised . In this way it maintains the pH of the solution .

Correct statements are as follows:

B) When a tiny amount of solvent is introduced to a buffer, the pH doesn't really alter considerably.

C) The ratio of acids to base concentration determines the pH of a phosphate buffer.

So, options B) and C) are correct.

ph scale used in science to measure the acidity or alkalinity of an aqueous solution. pH originally stood for "potential of hydrogen." The pH of acidic solutions is lower than that of basic or alkaline solutions.

Correct statements are as follows:

B) When a tiny amount of solvent is introduced to a buffer, the pH doesn't really alter considerably.

C) The ratio of acids to base concentration determines the pH of a phosphate buffer.

So, options B) and C) are correct.

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Answer:

C) A mass of 750 g and a volume of 70 dL .

Explanation:

Hello,

In this case, for substantiating the substance having the lowest density we need to compute it in the same units for each case as shown below:

[tex]\rho=\frac{m}{V}[/tex]

A) [tex]\rho =\frac{1.5kg}{1.2L}*\frac{1000g}{1kg} *\frac{1L}{1000mL}=1.25g/mL[/tex]

B) [tex]\rho =\frac{25g}{20mL}*=1.25g/mL[/tex]

C) [tex]\rho =\frac{750g}{70dL}*\frac{10dL}{1L}*\frac{1L}{1000mL} =0.107g/mL[/tex]

D) [tex]\rho =\frac{15mg}{25\mu L}*\frac{1g}{1000mg} *\frac{1000\mu L}{1mL}=0.6g/mL[/tex]

Therefore, the lowest density corresponds to C) A mass of 750 g and a volume of 70 dL

Regards.

Answer:

Mercury

Explanation:

Answer:

2,200 milligrams

Explanation:

40 mg per 1 kilogram means 40 times 55 which is 2,200 mg

Answer:

B - The amount of  OF2(g)  produced is halved.

Explanation:

Because its right.

When instead of 2 moles of F₂, 1  mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.

Let's consider the balanced reaction that occurs when F₂ reacts with NaOH.

2 F₂(g) + 2 NaOH(aq) → OF₂(g) + 2 NaF(aq) + H₂O(l)

First, let's see the moles of OF₂ and NaF obtained from 2 moles of F₂, using the molar ratios derived from the balanced chemical equation.

[tex]2 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 1 mol OF_2\\\\2 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 2 mol NaF[/tex]

Now, let's compare with the moles of OF₂ and NaF obtained from 1 mol of F₂, again using the same molar ratios derived from the balanced chemical equation.

[tex]1 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 0.5 mol OF_2\\\\1 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 1 mol NaF[/tex]

As we can see, since we have half the amount of F₂, we obtain half the amount of the products. Then, the only right option is: The amount of OF₂(g) produced is halved.

When instead of 2 moles of F₂, 1  mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.

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Answer:

No, it acts only as a reducing agent

Explanation:

Sulphur dioxide is a strong reducing agent. It is mainly used to prevent dried fruit from being oxidised and discoloured.

Answer: The 3rd and 6th bullet point is the quantitative data.

Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.  

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

N₂O₄ is the limiting reactant

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%