A chlorine concentration of 0.500 ppm is desired for water purification. What
mass of chlorine must be added to 2500 L of water to achieve this level?

Answers

Answer 1

Answer:

1.25 grams of chlorine.

Explanation:

Hello there!

In this case, since it is possible to define the parts per million of chlorine as the milligrams of chlorine per liters of water, in order to obtain the mass of chlorine in 2500 L, we proceed as follows:

[tex]ppm=\frac{mg}{L}\\\\mg=ppm*L[/tex]

In such a way, we plug in the given 0.500 ppm and 2500 L to obtain (in grams):

[tex]mg=0.500mg/L*2500L\\\\g=1250mg*\frac{1g}{1000mg}\\\\ g=1.25g[/tex]

Best regards!


Related Questions

In what type of reaction is water always a product

Answers

when hydrogen and oxygen combine then they form water as product

also in neutralization reaction water form as product

hope it helps

Water is always a product in a type of chemical reaction called a neutralization reaction.

A reaction is a process that involves the transformation of one or more substances into one or more different substances.

Neutralization reaction is type of reaction occurs when an acid and a base react together to form a salt and water. The general equation for a neutralization reaction is:

[tex]\rm acid + base \rightarrow salt + water[/tex]

In this type of reaction, the hydrogen ions ([tex]\rm H^+[/tex]) from the acid combine with the hydroxide ions ([tex]\rm OH^-[/tex]) from the base to form water ([tex]\rm H_2O[/tex]). The remaining ions from the acid and the base combine to form a salt.

Therefore, water is always a product in a neutralization reaction, which occurs when an acid and a base react together to form a salt and water.

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What pair of properties do all solids have?

Answers

Answer:

Has definite shape and volume

Solids have high melting point

Explanation:

How many moles are contained in 325 gram sample of ammonium dichromate (NH4)2Cr2O7

Answers

Answer:

Hyba odpowiedzi c. Hyba

Mole can be defined as the number of molar mass units of the compound in the sample. The moles of ammonium dichromate in 325 grams sample is 1.29 mol.

What is molar mass?

The molar mass can be given as the mass of each atom in the formula unit. The molar mass of ammonium chromate is given as:

[tex]\rm (NH_4)_2Cr_2O_7=2(N)+4\;\times\;2(H)+2(Cr)+7(O)[/tex]

Substituting the mass of each atom to identify the molar mass of ammonium chromate:

[tex]\rm (NH_4)_2Cr_2O_7=2(14)+4\;\times\;2(1)+2(52)+7(16)\\ (NH_4)_2Cr_2O_7=28+8+104+112\\ (NH_4)_2Cr_2O_7=252 \;g/mol[/tex]

The molar mass of ammonium dichromate s 252 g/mol. The moles of the compound in 325 g is given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass} \\\\Moles=\dfrac{325\;g}{252\;g/mol}\\\\ Moles=1.29\;mol[/tex]

The moles of ammonium dichromate in the sample are 1.29 mol. Thus, option c is correct.

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6. The drawing shows Samir riding his mountain bike.
i) Draw a circle around the places on the drawing where there
should be a lot of friction.
ii) Explain why there should be a lot of friction in these places.

Answers

Answer:

You Need To Upload The Photo Then I Will Help You :)

Explanation:

electronic configuration [Ne]3s2 3p6​

Answers

The answer is Argon! From Ne you go 2 3s level elements and then 6 3p level elements

The lithosphere ______ part of the earth’s crust.
A. Gaseous
B. Solid
C. Water
D. Liquid

Answers

Answer:

The lithosphere solid part of the earth's crust

I need help with chemistry :(

Answers

Answer:

How may we help kind sir

Explanation:

and if this was for points thanks

I don’t see the problem

Most cooking utensils are made up of aluminum because aluminum is _____​

Answers

Explanation:

Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.

because it is a lightweight material and a good heat conductor

By the reaction between
H3PO4 and Ca(OH)2,
Can be formede:

a. Ca(HPo4)2,
b. СaHP4
c. Ca(H2PO4)2
d. Ca2HP02
e.Сa3(PO4)2

the correct answers : b, c, e

the question is:
I am preparing for an admittion test in chemistry, I face this kind of questions which I dont how to understand or if I should only memorize.

I appreciate all your help.
thank you alot.​

Answers

Answer:

ANSWER is E

Explanation:

2H₃PO₄ + 3Ca(OH)₂→ 6H₂O + Ca₃(PO₄)₂

Answer:

First step is to find what the products would be. Since this seems to be a double displacement, you get: H3PO4(aq)+Ca(OH)2(aq)→ H2O(L)+Ca3(PO4)2(s)

You know the products because H has a charge of +1, so PO4 must have a charge of -3, and since OH has a charge of -1, Ca must be +2. Then make each compound have a net charge of 0. H20 is a liquid since aq solutions have liquid water. Ca3(PO4)2 is a solid, use the solubility rules.

Lastly, balance the equation: 2H3PO4(aq)+3Ca(OH)2(aq)→ 6H2O(L)+Ca3(PO4)2(s)

So the salt formed is Ca3(PO4)2

2 NaOH + H2SO4 → 2 H2O + Na2SO4
How many grams of water are produced from 203.50 grams of H2S04?

Answers

Answer:

[tex]m_{H_2O}=74.8gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to realize there is a 1:2 mole ratio of sulfuric acid to water; thus, given the mass of the former and its molar mass (98.07 g/mol), it is possible to determine the mass of produced water as shown below:

[tex]m_{H_2O}=203.50gH_2SO_4*\frac{1molH_2SO_4}{98.07gH_2SO_4}*\frac{2molH_2O}{1molH_2SO_4} *\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=74.8gH_2O[/tex]

Regards!

Which of the following components does a disc brake system use?
brake shoe
adjuster screw
wheel cylinder
caliper

Answers

Caliper I think I don’t really know but yea I think it’s caliper

The caliper is a component which is used by a disc brake system for performing its function.

What are the components of disc brake system?

The disc braking system consist of components such as disc/rotor, a brake calliper and brake pads. When the brake pedal is pushed, brake fluid creates pressure that produces friction.

So we can conclude that the caliper is a component which is used by a disc brake system for performing its function.

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4. Describe a simple experiment that can be performed in the laboratory to demonstrate
the formation of iron (III) chloride from iron fillings.​

Answers

Answer:

laboratory use the organic substances dissolve in water methanol or ethanol then Sinha to rise iron chloride solution is added a transient permanent coloration usually purple green or blue indicate the presence of a funnel or an hall

Balanced equation for Fe reacting with [tex]Cl_2[/tex]

[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]

What are iron fillings?

Iron filings are small shavings of ferromagnetic material.

When hydrochloric acid (HCl) reacts with iron filings (Fe) a dissolving metal reaction occurs with the Fe being converted to iron chloride ([tex]FeCl_2[/tex]) and hydrogen ([tex]H_2[/tex]).

The balanced equation is:

[tex]Fe + 2HCl[/tex] → [tex]FeCl_2 + H_2[/tex]

Note that reaction only produces [tex]FeCl_2[/tex] and not [tex]FeCl_3[/tex]. To make [tex]FeCl_3[/tex] you have to get more forcing and react Fe with chlorine gas ([tex]Cl_2[/tex]) at elevated temperature.

Since we are balancing equations here, below is the balanced equation for Fe reacting with [tex]Cl_2[/tex] (offered at no extra charge):

[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]

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Which of the following answers is true for the following statement?
The conjugate acid of a strong base is:
O aweak acid
O an amphiprotic acid
O a polyprotic acid
O a lousy (or very weak) acid
O a strong acid

Answers

it is either "aweak acid or a lousy (or very weak) acid"


What is the formula for the sulfate ion?

What is the formula for the sulfite ion?

What is the formula for the hydrogen sulfate ion?


Answers

Sulfate - SO4^-2

Sulfite - SO3^-2

Hydrogen Sulfate - HSO4-

Why is ocean water near the equator warmer than ocean water at the poles?
O A. The sun's rays strike the water more directly near the equator.
O B. Deep sea vents pump steam into the conveyor belt near the
equator.
O C. The water is denser and saltier near the equator, so it holds more
heat.
D. Due to Earth's rotation, gyres near the equator flow clockwise.
SUBM

Answers

I believe the answer is A. Hope this helps!

What volume, in mL, of 0.100 M NaOH is required to neutralize 35.0 mL of 0.102 M hydroiodic acid HI?

Answers

Answer:

35.7 mL NaOH

Explanation:

M1V1 = M2V2

M1 = 0.100 M NaOH

V1 = ?

M2 = 0.102 M HI

V2 = 35.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.102 M)(35.0 mL) / (0.100 M) = 35.7 mL NaOH

calculate the percentage of sodium and carbon and oxygen in sodium carbonate​

Answers

Percentage of Sodium [Na] in Sodium Carbonate [Na2CO3] = 46/106 x 100 = 43.40%. Percentage of Carbon [C] in Sodium Carbonate [Na2CO3] = 12/106 x 100 = 11.32%. Percentage of Oxygen [O] in Sodium Carbonate [Na2CO3] = 48/106 x 100 = 45.28% .....

Please help asap. Brainliest to correct.

Which formula can be used to calculate the percent yield? (5 points)

Select one:
a. (Theoretical yield ÷ actual yield) x 100

b. (Actual yield ÷ theoretical yield) x 100

c. (Amount of reactants ÷ amount of products) x 100

d. (Amount of products ÷ amount of reactants) x 100

Answers

Answer:

I guess this answer is c

yh

Answer:

b. (Actual yield ÷ theoretical yield) x 100

Explanation:

The percent yield will be a ratio of collected or actual yield to the theoretical yield. Like any percentage, you are determining a part of a whole. You multiply by 100 to convert your decimal answer from the division into a percentage.

what is the total of the amount of energy that is reflected back into space​

Answers

moon va half plus 22

3) How many grams of calcium hydroxide are necessary to create 49.5 mL of a 1.55 M solution?

Answers

Answer:

5.68g

Explanation:

What is the molar mass of Mg(HSO3)2?

Answers

186.47 is the correct answer I hope this helped!!!

Which statement best explains how weather is related to the water
cycle?

Answers

Answer:

Increased evaporationof water vapour from sea or land will increase rainfall

Answer:

Explanation:

increased evaporation of water vapor from sea to land will increase rainfall

A student is given an antacid tablet that weighs 5.4630 g. The tablet is crushed and 4.3620 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 13.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.7 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) has been neutralized in the 25.00 mL sample that was titrated

Answers

Solution :

It is given that :

Weight of the antacid tablet = 5.4630 g

4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.

25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.

27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]

                                                             = 12.27 ml of the original stomach acid.

What is the mass of 5.55 moles of carbon monoxide

Answers

Answer:

[tex]\boxed {\boxed {\sf 155 \ g\ CO}}[/tex]

Explanation:

To convert from moles to mass, the molar mass must be used.

First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.

Next, look up their molar masses on the Periodic Table.

C: 12.011 g/mol O: 15.999 g/mol

Since there is 1 atom of each, the molar masses can be added.

CO: 12.011 g/mol + 15.999 g/mol = 28.01 g/mol

Use this molar mass as a ratio.

[tex]\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

Multiply by the given number of moles:5.55

[tex]5.55 \ mol \ CO *\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

The moles of carbon monoxide cancel.

[tex]5.55 * \frac {28.01 \ g \ CO} {1 }[/tex]

Multiply.

[tex]155.4555 \ g \ CO[/tex]

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.

[tex]155 \ g\ CO[/tex]

5.55 moles of carbon monoxide is about 155 grams.

1 = 2 Amps, R = 3 Ohms, V = ? Volts​

Answers

Answer:

6 volts

Explanation:

Use Ohm's Law:

V = IR

V = (2 amps)(3 ohms) = 6 volts

how many grams of P3O9 contain 9.67x10^23 atoms of oxygen?​

Answers

Answer:

49.1 g

Explanation:

First we convert 9.67x10²³ atoms of oxygen into moles, using Avogadro's number:

9.67x10²³ atoms ÷ 6.023x10²³ atoms/mol = 1.60 mol O

Then we calculate how many P₃O₉ moles there would be with 1.60 O moles:

1.60 mol O * [tex]\frac{1molP_3O_9}{9molO}[/tex] = 0.18 mol P₃O₉

Finally we convert 0.18 P₃O₉ moles into grams, using its molar mass:

0.18 mol P₃O₉ * 237 g/mol = 49.1 g

A 9.56 sample of CuS is found to contain 6.35 g of copper what is the percent by mass of the copy and sulfur in this compound? (show your work)

Answers

Answer:

%Cu = 66.4%

%S = 33.6%

Explanation:

Step 1: Given data

Mass of CuS: 9.56 gMass of Cu: 6.35 g

Step 2: Calculate the mass of S

The mass of CuS is equal to the sum of the masses of Cu and S.

mCuS = mCu + mS

mS = mCuS - mCu

mS = 9.56 g - 6.35 g = 3.21 g

Step 3: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement / mCompound × 100%

%Cu = mCu / mCuS × 100% = 6.35 g / 9.56 g × 100% = 66.4%

%S = mS / mCuS × 100% = 3.21 g / 9.56 g × 100% = 33.6%

BRAINLIEST WILL BE GIVEN TO BEST ANSWERED!
What is the best explanation for why a magnet is different from a regular piece of metal?

A magnet has more electrons

The magnetic domains in a magnet are lined up with each other

Magnets are made of iron

A magnet has an electric field

Answers

Answer:

the second one

Explanation:

metal doesn't act like a magnet because the domains don't have a preferred direction of alignment. Magnets are all aligned in a specific direction.

Methane is combusted with Oxygen to produce Carbon Dioxide and Water. How many grams of oxygen are required to react with 55 grams of
Methane?
A.55 g
B.110 g
C.165 g
D.220 g

Answers

The answer is B.110 g

If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to kinetic energy?

Answers

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength [tex]g[/tex] is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If [tex]m[/tex] denote the mass of this ball, the gravitational potential energy ([tex]\rm GPE[/tex]) of this ball at height [tex]h[/tex] would be [tex]{\rm GPE} = (m \cdot g) \cdot h[/tex], which is proportional to [tex]h\![/tex].

The value of [tex]g[/tex] near the surface of the earth is indeed approximately constant (typically [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex].)

At halfway between the top and bottom of this hill, the height of this ball would be [tex](1/2)[/tex] of its initial value (the value when the ball was at the top of the hill.) Because the [tex]\rm GPE[/tex] of this ball is proportional to its height, at halfway down the hill, the [tex]\rm GPE\![/tex] of this ball would also be [tex](1/2)\![/tex] its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that [tex](1/2)[/tex] of the initial [tex]\rm GPE\![/tex] of this ball was not lost. Rather, these [tex](1/2)\![/tex] of the initial [tex]\rm GPE[/tex] would have been converted to the kinetic energy ([tex]\rm KE[/tex]) of this ball.

Hence, when the ball is halfway down the hill:

[tex]\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}[/tex].

[tex]\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}[/tex].

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

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