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BARSIC [14]
3 years ago
6

The largest unit of geologic time in the list below is:​

Chemistry
1 answer:
sergey [27]3 years ago
3 0

Answer:

From largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages. All of these are displayed in the portion of the geologic time scale shown below. The Phanerozoic Eon represents the time during which the majority of macroscopic organisms — algae, fungi, plants and animals — lived.

Explanation:

Hope this helps you :)

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How many moles are in 18.2 g of CO2?<br> 41.4 moles<br> 801 moles<br> 0.414 moles<br> 0 2.42 moles
mrs_skeptik [129]

Answer:

0.414 mole (3 sig. figs.)

Explanation:

Given grams, moles = mass/formula weight

moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)

≅ 0.414 mole (3 sig. figs.)

6 0
2 years ago
a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRT

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0
3 years ago
If 5.0 g of copper metal reacts with a solution of silver nitrate, how many grams of silver metal are recovered?
sukhopar [10]

Answer:

16.9g

Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2  

Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.

Then to calculate the theoretical yield you need to compare moles of the reactants:

m(Cu)=5g

M(Cu)=63.55

n(Cu)=5/63.55=0.0787

By comparing coefficients you require twice as much silver: 0.157mol

n(Ag)=0.157

M(Ag)=107.86

m(Ag)=0.157x107.86=16.9g

Hence, the theoretical yield of this reaction would be 16.9g

3 0
2 years ago
The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re
Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

\log D=\log D_o-\frac{E_a}{2.303RT}

where,

D = diffusion coefficient = 2.7\times 10^{-14}m^2/s

D_o = pre-exponential constant = 1.7\times 10^{-5}m^2/s

E_a = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K

Hence, the temperature of the system will be 1622 K

8 0
3 years ago
To see if it works like they want it to, engineers need to Question 3 options:
ziro4ka [17]

Answer:

they need to do their research

4 0
2 years ago
Read 2 more answers
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