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Citrus2011 [14]
3 years ago
6

HELP PLEESE HELP.....​

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

i'm sorry

Explanation:

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What is the molar mass for FeBr3
AleksAgata [21]
Mass of Iron (Fe): 55.845Mass of Bromine (Br): 79.904
You need to multiply the mass of Br by 3 because there are 3 Bromine atoms. 
(79.904)(3)+ 55.845= 239.712+55.845                                = 295.557 g/mol

6 0
3 years ago
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.
timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU +  w

q = 475 J  +  93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0
3 years ago
8. 00 g of ethane gas, c2h6, is burned in oxygen. What volume of carbon dioxide gas is produced at 1. 00 atm and 25. 0°c?.
Sonbull [250]

Answer:

Explanation:1 g CH

4

​

,C

2

​

H

6

​

C

3

​

H

8

​

C

4

​

H

10

​

T = 350 K, P = 1 atm

PV=nRT=

M

w×R×T

​

1×V=

58

1

​

×0.0821×350

V=0.495 L

V=495cm

3

8 0
1 year ago
How would your atomic mass prediction have been affected if not all of the water had been evaporated from the calcium carbonate
Yanka [14]
Go to a famlie member, if you can't trust them, trust god. if you cant trust god, BURN IN HELL
6 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
3 years ago
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