All categories

2 years ago

Can someone plz help me? :(

Chemistry

1 answer:

Tju [1.3M]2 years ago

8 0

Answer:

the Respiratory System and Circulatory system

Explanation:

You might be interested in

Indica, con base en la siguiente ecuación:

Anvisha [2.4K]

Hello I just want to say have a good day

7 0

2 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

2 years ago

How is the use of word theory in science different from its more common use

weqwewe [10]

It is like an idea that they come up with in an experimeant

4 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

2 years ago

Identify and explain exothermic and endothermic reactions in terms of energy transfer

Ulleksa [173]

Exothermic reaction is when the system releases energy to the surrounding, the energy (in the perspective of the system) will have a sign of negative, as the system is transferring energy towards the surrounding.

Endothermic reaction is when the system is absorbing energy from the surrounding, the energy (in the perspective of the system) will have a sign of positive, as energy is transferred to the system.

5 0

3 years ago