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Nadusha1986 [10]
3 years ago
15

Can someone help me with #35,37,39,41 please

Chemistry
1 answer:
adoni [48]3 years ago
4 0
35 b.
37 h
39 i
41 d
hope that helps
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Enter your answer in the provided box.
nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, V_1=43.1\ L

Initial temperature, T_1=24^{\circ} C=24+273=297\ K

Final temperature, T_2=78^{\circ} C=78+273=351\ K

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L

So, the final volume of the gas is 50.93 L.

4 0
3 years ago
According to the graph, which statement is true?
damaskus [11]
Nitrogen is the one
5 0
3 years ago
State the importance of oxygen for germination process
tamaranim1 [39]
Plants need oxygen to survive. 
3 0
3 years ago
Which of the following symbols represents a chlorine ion with a stable arrangement of eight valence electrons?
frozen [14]

A stable arrangement of eight valence electrons : ³⁵Cl⁻¹

<h3>Further explanation</h3>

Chlorine is a halogen gas, located in group 17, p block

Chlorine has an atomic number of 17 and an atomic mass of 35

Electron configuration: [Ne] 3s²3p⁵

If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)

So by binding this one electron, chlorine forms negative ions (anions)

³⁵Cl⁻¹

B. Cl⁻² binds 2 electrons, exceeding the octet rule

C. Cl⁺¹, releases 1 electron, remains unstable

D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration

3 0
3 years ago
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal
bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = \frac{0.840}{148.2}moles of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release (5539\times 0.00567)kJ of heat or 31.41 kJ of heat

6.60 kJ of heat increases 1^{0}\textrm{C} temperature of calorimeter.

So, 31.41 kJ of heat increases (\frac{1}{6.60}\times 31.41)^{0}\textrm{C} or 4.76^{0}\textrm{C} temperature of calorimeter

So, the final temperature of calorimeter = (20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}

3 0
3 years ago
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