All categories

3 years ago

What three things were a result of the fall or Rome

Chemistry

1 answer:

juin [17]3 years ago

6 0

Answer:

The politicians and rulers of Rome became more and more corrupt.
Infighting and civil wars within the Empire.

Attacks from barbarian tribes outside of the empire such as the Visigoths, Huns, Franks, and Vandals.

You might be interested in

How many moles of NaCl are contained in 50.0 mL of 2.50 M NaCl?

GarryVolchara [31]

Answer: 0.144 moles. :)

Explanation:

5 0

3 years ago

What is the molecular formula for Tricarbon nonachloride?

vampirchik [111]

Answer:

$C_{3}Cl_{9}$

Explanation:

Molecular Formula is representation of the chemical compound in terms of the symbols of all the elements that are present in the compound followed by subscripts, which give the count of each element in that compound.

We need to write the molecular formula of Tricarbon nonachloride. Tri means three, so Tricarbon means there are 3 atoms of Carbon. Likewise, nona stands for 9, so nonachloride means there are 9 atoms of chlorine. Therefore, we can represent nonachloride as:

Carbon (3 atoms) Chlorine (9 atoms) = $C_{3}Cl_{9}$

Thus, molecular formula of Tricarbon nonachloride is $C_{3}Cl_{9}$

8 0

3 years ago

Read 3 more answers

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

3 years ago

What is the name of the molecule below

il63 [147K]

Answer:

2-octene

Explanation:

the 8-carbon chain (oct); has a double bond (ene) is on the 2nd carbon (2-)

4 0

3 years ago

The molar mass of argon is 40 g/mol. what is the molar mass of a gas if it effuses at 0.91 times the speed of argon gas?

Gemiola [76]

To determine the molar mass of the unknown gas, we use Graham's Law of Effusion where it relates the effusion rates of two gases with their molar masses. It is expressed as r1/r2 = √M2/M1. We calculate as follows:
Let 1 = argon gas 2 = unknown gas
r2 = 0.91r1r1/r2 = 1/0.91
1/0.91 = √M2/M1 = √M2/40M2 = 48.30 g/mol

8 0

4 years ago

Read 2 more answers