an engineering team is conducting a trial launch of a few rocket . which part is the engineering process is the team in ?

Answer: sorry I cant help you I need the same answer

Explanation:

Answer:

b

Explanation:

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.

Learn more about:

Answer:

compound and elements.

Explanation:

Answer:

Electrolytes are minerals in your body that have an electric charge. They are in your blood, urine, tissues, and other body fluids. Electrolytes are important because they help. Balance the amount of water in your body

Explanation:

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:

1007.94 J

Explanation:

Work done: This can be defined as the product of force and distance, along the direction of the force. The S.I unit of work is Joules (J) .

Form the question above,

W = F×d........................... Equation 1

Where W = Work done, F = Resistive force, d = distance.

Given: F = 21.4 N, d = 47.1 m

Substitute these values into equation 1

W = 21.4(47.1)

W = 1007.94 J

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:

17.5609grams

Explanation:

According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams i.e.

= 6.814 + 0.08753

= 6.90153 grams

The result of this mixture i.e. 6.90153 grams is divided into three equal parts i.e.

= 6.90153 ÷ 3

= 2.30051 grams

This means that each equal part weighs 2.30051 grams

One of the yield (2.30051 grams) is multiplied by 7.6335 times i.e.

= 2.30051 × 7.6335

= 17.56094309

Hence, the final mass is 17.5609grams

Answer:

2,200 milligrams

Explanation:

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

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]

Answer:

A. Neopentane.

B. 2,2-dimethyl-propane.

Explanation:

Hello,

In this case, for the given compound we can spell out two possible names:

A. Neopentane: it is, because the molecule has five carbon atoms wherein there is a central one, bonded to the other four.

B. 2,2-dimethyl-propane: it is, because we find two methyl radicals at the second carbon.

Regards.

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: The correct answer would be : True

I hope that this helps you !

Answer:

Mercury

Explanation:

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%

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: 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: Yes

Explanation:

Density of a liquid depend on its volume. This is because Density is mass of liquid divided by volume.

Density is inversely proportional to volume.

As density increases, volume decreases and vice versa. The density for water is 1g/ milliliter but it changes with changes in temperature or there are impurities dissolved in it. Ice is less dense that liquid water and it's the major reason it's float because it's volume is inversely proportional to it's density.

want to talk

Explanation:

Answer:

D.ionic interaction is a strong interaction

When atoms come close enough to one another, an ionic connection is created. Atoms can interact with one another ionically, and this distance is greater than the bonding distance. Therefore, option D is correct.

The main interaction taking place in ionic compounds is ionic bonding, a form of chemical bonding that includes the electrostatic attraction between two atoms or ions with dramatically differing electronegativities.

Ions with opposing charges are drawn to one another through ion-ion interactions. They are also known as ionic bonds, because they are what keep ionic compounds together. opposing charges attract each other whereas like charges repel each other.

The whole transfer of valence electrons between atoms is referred to as an ionic bond. It is a kind of chemical connection that produces two ions with opposing charges. In ionic bonding, the nonmetal receives the electrons to become a negatively charged anion while the metal gives them up to become a positively charged cation.

Thus, option D is correct.

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#SPJ5

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 .

Answer: the two parameters change inversely; as hydrogen ion concentration increases, pH falls. due to the logarithmic relationship, a large change in hydrogen ion concentration is actually a small change in pH. For example, doubling the hydrogen ion concentration causes pH to fall by just 0.3.

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:

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:

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: 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:

[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:

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!