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pav-90 [236]
2 years ago
7

A 20.0-ml sample of glucose Iv solution has a mass of 20.6g.what is the density of the glucose​

Chemistry
1 answer:
xxTIMURxx [149]2 years ago
4 0

Taking into account the definition of density, the density of the glucose is 1.03 \frac{g}{mL}.

<h3>What is density</h3>

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Density of the glucose</h3>

In this case, you know that:

  • Density= ?
  • Mass= 20.6 g
  • Volume= 20 mL

Replacing in the definition of density:

density=\frac{20.6 g}{20 mL}

Solving:

<u><em>density= 1.03 </em></u>\frac{g}{mL}

In summary, the density of the glucose is 1.03 \frac{g}{mL}.

Learn more about density:

brainly.com/question/952755

brainly.com/question/1462554

#SPJ1

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Some chemical heat highly temperature it was produce poison gas
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3 years ago
Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid
jasenka [17]

Answer: a) The K_a of acetic acid at 25^0C is 1.82\times 10^{-5}

b) The percent dissociation for the solution is 4.27\times 10^{-3}

Explanation:

CH_3COOH\rightarrow CH_3COO^-H^+

 cM              0             0

c-c\alpha        c\alpha          c\alpha

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.10 M and \alpha = ?

Also pH=-log[H^+]

2.87=-log[H^+]  

[H^+]=1.35\times 10^{-3}M

[CH_3COO^-]=1.35\times 10^{-3}M

[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M

Putting in the values we get:

K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}

K_a=1.82\times 10^{-5}

b)  \alpha=\sqrt\frac{K_a}{c}

\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}

\alpha=4.27\times 10^{-5}

\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}

5 0
3 years ago
Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so
dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

<h3>3.18 (w/w) % </h3>
4 0
3 years ago
Increase the mass of the red ball. What happens to the force on the blue ball? What happens to the force on
gregori [183]

Answer:

i really dont know because i dont have anything to look at to know that answer.

5 0
3 years ago
If 185 g of KBr are dissolved in 1.2 kg of water, what would be the expected change in boiling point? The boiling point constant
Fudgin [204]

Answer:

ΔTb = 0.66 C

Explanation:

Given

Mass of KBr = 185 g

Mass of water = 1.2 kg

Kb = 0.51 C/m

Explanation:

The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)

\Delta T_{b}= K_{b}* m

Molality = \frac{moles\ KBr}{Kg\ water} \\\\moles KBr = \frac{mass\ KBr}{Mol.wt\ KBr} = \frac{185}{119} = 1.555\\\\Molality (m) = \frac{1.555 }{1.2} =1.296 m\\

[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]

6 0
3 years ago
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