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tigry1 [53]
3 years ago
7

A sample of neon gas is at a temperature of 60°C and has a volume of 300 cm. What is the volume of the gas if it is heated to 10

0°C at a constant pressure?​
Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Answer:

336.04 cm

Explanation:

T1 = 60+273 = 333K

T2 = 100+273 = 373K

V2 = (V1×T2)/T1

V2 = 300×373/333

V2 = 336.04 cm

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How could you engineer a device that produces beneficial friction and heat?
nlexa [21]

Answer:

Hi

Explanation:

That's why rubbing your hands together makes them warmer. ... Friction causes the molecules on rubbing surfaces to move faster, so they have more energy. This gives them a higher temperature, and they feel warmer. Heat from friction can be useful.

3 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
4Na + O2 2Na2O
beks73 [17]
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98

g
⋅
m
o
l
−
1
. If
5

m
o
l
natrium react, then
5
2

m
o
l
×
61.98

g
⋅
m
o
l
−
1

=

154.95

g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go
5 0
3 years ago
Use the periodic table to write the electron configuration for rubidium (Rb) in noble has notation
Alexxx [7]

Answer:

Rb: [Kr] 5s  

Step-by-step explanation:

Rb is element 37, the first element in Period 5.

It has one valence electron, so its valence electron configuration is 5s.

The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.

The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.

4 0
3 years ago
How do river systems, watersheds, and divides interact?
love history [14]

Answer:

They all end up in the oceon

Explanation:

5 0
3 years ago
Read 2 more answers
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