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marshall27 [118]
3 years ago
15

Why does your ankle feels cold when you apply a cold pack to it?

Chemistry
1 answer:
Tresset [83]3 years ago
5 0

Answer:

a.) because your ankle is taking in cold energy from the cold pack

Explanation:

all of the other ones say heat, which is not applied in this question

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An alpha particle is identical to a helium atom that has been stripped of its two electrons
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Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).
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Oxen beacon is great for your self halllway
7 0
3 years ago
What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO
docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

  • \rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0
3 years ago
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0
3 years ago
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

Therefore, the mass of PbSO_4 needed are, 1.515 grams.

6 0
3 years ago
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