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Anarel [89]
3 years ago
11

This is a dumb question but does smelling shapies make you lose brain cells?

Chemistry
2 answers:
hammer [34]3 years ago
7 0

Yes````````````````````````````````````````

nadezda [96]3 years ago
5 0

Yes, Sharpies contain volatile solvents—and when inhaled these solvents can produce a "high." The effects of inhalants (including Sharpies) can be similar to those of alcohol and include slurred speech, lack of coordination, euphoria, and dizziness.

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What are the products of the following equation?
Artyom0805 [142]
The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
5 0
2 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
Both E. coli and Salmonella are single-celled organisms. They do not have a nucleus or other membrane-bound organelles. Based on
AfilCa [17]

Answer:

domain bacteria

Explanation:

Salmonella and E. coli are same in the sense that they are both bacteria,

7 0
3 years ago
consider the titration of hclo4 with koh. what is the ph after 17.0 ml of 0.15 m koh has been added to 15 ml of 0.20 m hclo4?
bezimeni [28]

The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is  <u>3.347</u>.

Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.

<u>Calculation:-</u>

Normality of acid                                               Normality of base

= nMV                                                                        nMV

= 1 × 0. 15 × 0.017                                              1 ×  0. 20 ×0.015 L

= 2.55 × 10⁻³                                                             = 3 × 10⁻³

The overall base will be high

net concentration = 3× 10⁻³ - 2.55 × 10⁻³

                             = 0.45 × 10⁻³

                             = 4.5× 10⁻⁴

pH = -log[4.5 × 10⁻⁴]

    = 4 - log4.4

     = <u>3.347</u>

A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.

Learn more about titration here:-brainly.com/question/186765

#SPJ4

7 0
1 year ago
Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit
Aleksandr-060686 [28]

Answer:

+ axit

CH2=CH-CH2-COOH,

CH3-CH=CH-COOH (tính cả đồng phân hình học)

CH2=C(CH3)-COOH.

+ este

HCOOCH=CH-CH3 (tính cả đồng phân hình học)

HCOO-CH2-CH=CH2,

HCOOC(CH3)=CH2.

CH3COOCH=CH2

CH2=CH-COOCH3

8 0
3 years ago
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