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borishaifa [10]
3 years ago
8

Calculate 139 centimeters cubed argon at STP

Chemistry
1 answer:
rewona [7]3 years ago
3 0
Volume in liters:

139 cm³ / 1000 = 0.139 L   

molar mass Argon = 39.95 g/mol

1 mole -------- 22.4 L ( at STP )
? mole ------- 0.139 L

moles of argon :

0.139 * 1 / 22.4 = 0.00620 moles 

Mass of argon :

0.00620 * 39.95 = 0.24769 g

hope this helps!
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<span>The test dummy will continue forward until it makes contact with another object.</span>
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Bumek [7]

Answer:

1 mole of C2H6.

Explanation:

The balanced equation for the reaction is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

We can determine the number of mole of C2H6 that reacted to produce 2 moles of CO2 as follow:

From the balanced equation above,

2 moles of C2H6 reacted to produce 4 moles of CO2.

Therefore, Xmol of C2H6 will react to produce 2 moles of CO2 i.e

Xmol of CO2 = (2 x 2)/4

Xmol of CO2 = 1 mole.

Therefore, 1 mole of C2H6 is required to produce 2 moles of CO2.

6 0
3 years ago
Compare the chemical reactivity of chlorine and bromine. Explain your answer. Thanks in advance
shepuryov [24]

Bromine vs Chlorine | Br vs Cl

 

Halogens are group VII elements in the periodic table, and all are electronegative elements and have the capability to produce -1 anions.

Bromine

Bromine is denoted by the symbol Br. This is in the 4th period of the periodic table between chlorine and iodine halogens. Its electronic configuration is [Ar] 4s2 3d10 4p5. The atomic number of bromine is 35. Its atomic mass is 79.904. Bromine staChlorine is an element in the periodic table which is denoted by Cl.  It is a halogen (17th group) in the 3rd period of the periodic table. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Its electron configuration is written as 1s2 2s2 2p6 3s2 3p5. Since the p sub level should have 6 electrons to obtain the Argon, noble gas electron configuration, chlorine has the ability to attract an electron. ys as a red-brown color liquid at room temperature.

8 0
3 years ago
What is the density of an 800 gram object that occupies 200 cm 3 ?
vova2212 [387]
Density = mass/volume = 800/200 =4 g/cm^3
8 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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