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3 years ago

Bicarbonate concentrate mixers may have a which are replaced on a routine basic.

Chemistry

1 answer:

Sidana [21]3 years ago

3 0

Answer: True the bicarbonate mixture can help save time and few routine.

Explanation:

For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.

Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods.

A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.

Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.

Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.

The answer to the above is true the bicarbonate mixture can help save time and few routine.

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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago

PLEASE PEOPLE I NEED HELP PLEASE IF YOU SEE THIS HELP ME ! :,( !!!!! PLEASEEEEEEEEE

Tems11 [23]

Answer:

year 1 is 5.5%

year 2 is 7.5%

year 3 is 10.2%

Explanation:

since,

length of the transect covered in seaweed / total lenth of transect x 100

then,

0.55 / 10.0 x 100 = 5.5

and

0.75 / 10.0 x 100 = 7.5

and

1.02 / 10.0 x 100 = 10.2

you could also just move the decimal to the right once

4 0

2 years ago

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c

Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

6 0

3 years ago

I need help.connect now

olya-2409 [2.1K]

And the help is.....?!

4 0

3 years ago

Read 2 more answers

The discovery of the atomic nucleus, or perhaps better stated the development of theory of the nucleus, is credited to Ernest Ru

Greeley [361]

Answer:

Both of the studies said that the mass of the atom is centered in the nucleus, which is positive, and there are electrons (negative particles) orbiting it. So, Rutheford and Nagaoka discovered that the atom can be divisible and it has an empty space.

But, in the model of Nagaoka, the nucleus was huge, and for Rutherford, the nucleus was really small, and the mass was concentrated. By his experiment with the gold sheets, the theory was appropriated. That's why Rutherford is credited with the discovery of the nucleus. Nagaoka was incorrect in his suppositions.

3 0

3 years ago