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ArbitrLikvidat [17]
2 years ago
5

A liquid with high viscosity _____ flow easily and _____ effective in wetting a surface.

Chemistry
1 answer:
scoundrel [369]2 years ago
5 0

A liquid with high  viscosity does not flow easily and is not effective in wetting a surface.

When a metal is subjected to corrosive elements including salt, moisture, and high temperatures, a reaction called corrosion takes place inside the metal. Some foods contain metallic compounds that can corrode a material. The majority of corrosion is simply surface dis-colouration, which polishing agents may quickly remove.

Increasing viscosity and constant intermolecular water bonding together result in surface tension. Any liquid that was more viscous than water possessed a surface tension that was equal to or lower than that of water. Viscosity with surface tension decrease when temperature rises.

Therefore, a liquid with high viscosity does not flow easily and is not effective in wetting a surface.

To know more about viscosity

brainly.com/question/2193315

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Which of these forces is acting on a baseball sitting on level ground? A. applied force B. frictional force C. gravitational for
snow_lady [41]

C. gravitational force

4 0
3 years ago
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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
15.0 g of Fe and 25.0 g of sand are added to 250.0 g of water. a. Determine the percent mass of Fe, sand, and water in the mixtu
lana [24]

Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand  = 8.62%

percentage mass of water = 86.205%

B. (Iron + sand + water) -------> ( iron + sand) ------> sand

C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

percentage mass of iron = 15/290 * 100% = 5.17%

percentage mass of sand = 25/290 * 100% = 8.62%

percentage mass of water = 250/290 * 100% = 86.205%

B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

5 0
3 years ago
FIRST GETS BRAINLIEST
lisov135 [29]

Answer:

0.07172 L = 7.172 mL.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT. </em>

where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K, Standard T).

<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>

8 0
3 years ago
The gas usually filled in the electric bulb is
Mariulka [41]

Answer:

1st answer:  A. nitrogen

2nd answer:  A. Sodium carbonate

Explanation:

8 0
3 years ago
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