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Serggg [28]
2 years ago
8

What is the molarity of na+ in a solution of nacl whose salinity is 5.6 if the solution has a density of 1.03 g/ml?

Chemistry
1 answer:
pantera1 [17]2 years ago
8 0
Salinity has units of grams NaCl or salt per kilogram solution. We can use the density given and the molar mass of the salt to convert from salinity to molarity. We do as follows:

( 5.6 g / kg ) ( 1.03 kg / L ) ( 1 mol / 58.44 g ) = 0.0987 mol NaCl / L
You might be interested in
If 50% of a radioactive element remains after 4000 years, what is the half-life?
ANTONII [103]

Hey there!

A half-life means after a certain amount of time, half of that substance will be gone/changed after that time.

If 50%, or half, of the element remains after 4000 years, that means the half life must be 4000 years.

Hope this helps!

4 0
3 years ago
CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
nirvana33 [79]

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

3 0
3 years ago
Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydr
vlabodo [156]

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2  ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5


4 0
2 years ago
Read 2 more answers
You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans
OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

4 0
3 years ago
What is the most important requirement for all living things? water carbon dioxide light vitamins
Sidana [21]
Water. All organisms that depend on oxygen need water to live.
5 0
3 years ago
Read 2 more answers
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