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3 years ago

Is drying wet clothes a physical or chemical change

2 answers:

8 0

Drying clothes is a physical change, because it is not permanent. The clothes could just be wettened again, therefore it is not a chemical change.

~Hope I helped!~

Nadya [2.5K]3 years ago

5 0

Drying wet clothes would be seen as a physical change. This is because physical changes are changes that affect the form of a chemical substance, but not its chemical composition. Drying wet clothes did not change its chemical composition, therefore the answer to this question IS in fact, a physical change!
(hope it helped ^_^)

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Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -

marusya05 [52]

Answer:

= 2.2 g pf S. produced

Explanation:

Balanced Reaction equation:

$2H_{2} S + SO_{2}$ → $3S + 2H_{2} O$

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 = 0.09 moles of H2S

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 = 0.069 moles of SO2

Meaning SO2 is the limiting reagent

Finally, 3 moles of S - 32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

= 2.2 g pf S.

7 0

3 years ago

What are the concentrations of h3o+ and oh− in tomatoes that have a ph of 4.10?

IRINA_888 [86]

Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

pH : pH is defined as the negative logarithm of hydronium ion concentration.

Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

$4.10=-log[H_3O^+]$

$[H_3O^+]=antilog(-4.10)$

$[H_3O^+]=7.94\times 10^{-5}$

Now we have to calculate the pOH.

As we know, $pH+pOH=14$

$4.10+pOH=14$

$pOH=9.9$

Now we have to calculate the hydroxide ion concentration.

$pOH=-log[OH^-]$

$9.9=-log[OH^-]$

$[OH^-]=antilog(-9.9)$

$[OH^-]=1.258\times 10^{-10}$

Therefore, the concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

5 0

3 years ago

Why do the densities of most substances increase when temperature decreases?

vlada-n [284]

This is because temperatures determine the kinetic energy of molecules of a substance, At lower temperatures the molecules have low kinetic energy hence the distance between molecules is not as large as when the kinetic energy is higher (because the molecules bombard less and with less kinetic energy). This means the substance can pack more molecules per volume at lower temperatures. The more the molecules per volume the higher the density.

6 0

3 years ago

Read 2 more answers

CHEM PLZ HELP

marusya05 [52]

Answer:

Increase in CO2 (g) over time.

No NaHCO3 (s) will be left after a time

Explanation:

The reaction, shown below;

2NaHCO3(s) → Na2CO3(s)+CO2(g)+H2O(ℓ) is a decomposition reaction. A decomposition reaction is a kind of chemical reaction in which a given chemical specie breaks up to give other chemical species. Decomposition may be induced by heat or light.

Usually, there is only one reactant in a decomposition reaction; the specie that disintegrates into the products. This reactant usually decreases in concentration steadily because it is converted into products. This is why the mass of NaHCO3(s) in the system continues to decrease steadily until it finally falls to zero.

Conversely, the concentration (for aqueous) or volume (for gases) or mass (for solid) products of the reaction increases steadily as the reaction progresses. This explains why the volume of CO2 in the system will steadily increase over time.

7 0

3 years ago

Object A specific heat is 2.45 J/g C and object B specific heat is 0.82 J/g C. Which object will heat up more if they have the s

Artist 52 [7]

Answer: Object B will heat up more.

Explanation:

The formula for specific heat is as follows.

Q = $m C \Delta T$

Where,

Q = heat provided

m = mass

C = specific heat

$\Delta T$ = change in temperature

Now, both the objects have same mass and equal amount of heat is applied.

According to the formula, the equation will be as follows.

$Q_{1}$ = $Q_{2}$

$m C_{1}\Delta T$ = $m C_{2}\Delta T_{2}$

Cancel m from both sides, as mass is same. Therefore,

$C_{1} (T_{f} - T_{i})_A$ = $C_{2}(T_{f} - T_{i})_B$

Cancel out the initial temperature and put the values of specific heat, then the equation will be as follows.

$2.45 (T_{f})_A$ = $0.82(T_{f})_B$

Therefore, from the above equation it can be concluded that the object with low specific heat will heat up more as its specific heat will be inversely proportional to its final temperature.

Hence, object B will heat up more.

4 0

3 years ago