In what state of matter are the particles vibrating at high speeds?

What are the relationships between temperature and viscosity of water?

iris [78.8K]

Explanation:

Both cohesion and molecular interchange contribute to liquid viscosity. The impact of increasing the temperature of a liquid is to reduce the cohesive forces while simultaneously increasing the rate of molecular interchange. The former effect causes a decrease in the shear stress while the latter causes it to increase.

temperature?

The viscosity of liquids decreases rapidly with an increase in temperature, and the viscosity of gases increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas gases flow more sluggishly.

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8 0

3 years ago

A 3.42 gram sample of an unknown gas is found to occupy a volume of 1.90 L at a pressure of 547 mm Hg and a temperature of 33 °C

liq [111]

Use the Ideal Gas Law to find the moles of gas first.

Be sure to convert T from Celsius to Kelvin by adding 273.

Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.

PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K

n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas

Now divide grams by mol to get the molecular weight.

3.42 g / 0.0545 mol = 62.8 g/mol

3 0

3 years ago

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %

Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol$