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ale4655 [162]
3 years ago
12

What does burning bread in a toaster demonstrate? (a) Energy can create a change. (b) Energy can cause motion. (c) A toaster cre

ates its own energy. (d) A toaster is needed to burn bread.
Chemistry
1 answer:
ra1l [238]3 years ago
3 0

Answer:

A option is correct.

(a) Energy can create a change.

Explanation:

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Create: Click Reset. Use the Gizmo to create a path in which the carbon atom goes from the atmosphere to the hydrosphere, biosph
Solnce55 [7]

Answer:

From the atmosphere to the hydrosphere by diffusion.

From the atmosphere to the biosphere by photosynthesis.

From the atmosphere to the geosphere by rainfall.

Explanation:

Carbon atom goes from the atmosphere to the hydrosphere by the process of diffusion because there is high concentration of carbondioxide present in the atmosphere. The carbon atom goes from the atmosphere to the biosphere by the process of photosynthesis in plants which uses carbondioxide gas as a raw material in the process for the preparation of organic compounds such as glucose. The carbon atom goes from the atmosphere to the geosphere with the help of rain. When carbondioxide gas react with water in the atmosphere, carbonic acid is formed and comes to the ground through rainfall.

7 0
3 years ago
Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

6 0
3 years ago
Read 2 more answers
How many carbon atoms are in 26-hydrogen alkyne
navik [9.2K]
An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)
5 0
3 years ago
If the H3O is 4.950 x 10-12 what is the ph?
romanna [79]

Answer:

pH = 11.3

Explanation:

From the question given above, the following data were obtained:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

The pH of a solution is defined by the following equation:

pH = –Log [H₃O⁺]

Thus, with the above formula, we can obtain the pH of the solution as follow:

Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M

pH =.?

pH = –Log [H₃O⁺]

pH = –Log 4.950×10¯¹²

pH = 11.3

8 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
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