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3 years ago

Mendeleev created the first periodic table by arranging the elements in the order of ?

Chemistry

1 answer:

Maslowich3 years ago

4 0

Mendleev arranged his periodic table in order of Atomic Mass...

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When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu

olga2289 [7]

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC = C = 0,50 J/gºC

8 0

3 years ago

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0

3 years ago

19)Calculate the number of moles of Al2O3 that are produced when 15 mol of Fe is produced

Margarita [4]

Answer:

Explanation:

Calculate the number of moles of Al2O3 that are produced when 15 mol of Fe is produced

in the following reaction.

C) 45 mol

6 0

2 years ago

Calculate the density of carbon dioxide at STP

Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

5 0

3 years ago

Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?

34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0

3 years ago