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tatuchka [14]
3 years ago
12

Write the electron dot structures for the following elements:

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

Drawings of Lewis structure Are attached for your elements

Explanation:

Please open the attachment

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Which of the following statements accurately describes enzymes?
aniked [119]
The answer is a. temperature and pH can affect how the enzymes work
5 0
3 years ago
You titrate 25.00 mL of the lemon-lime Kool-Aid (with KI, HCl, and starch) with 0.001000 M KIO3(aq) solution. The titration requ
Svetradugi [14.3K]

Answer:

1.0190 x 10⁻⁵ mol

Explanation:

We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).

Molarity = mol/V

V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L

⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L =  1.0190 x 10⁻⁵ mol KIO₃

# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol

8 0
3 years ago
A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L
Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL

So, the new volume is 150 mL.

8 0
3 years ago
Which process is an example of a physical change?
Lostsunrise [7]

Answer:

The answer is flattening

Explanation:

A physical change is generally something that affects the shape of form of the matter and a chemical change results from a chemical reaction. Flames are caused by chemical reactions, as is rust, and the process of a fruit becoming ripe. Thus, the answer is “flattening”.

4 0
3 years ago
describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid​
Reptile [31]

Answer:

Explanation:

250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.

150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.

5 0
3 years ago
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