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3 years ago

All gases are considered to be ____.

Chemistry

1 answer:

NISA [10]3 years ago

3 0

Answer:

Explanation:

I used my notes from class today.

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Help help help plsss

Olin [163]

Answer:

4.58×10^3

3.72×10^7

Explanation:

3.8×10^2 =380

+ 4.2×10^3=4200

= 4580 = 4.58×10^3

8 0

2 years ago

Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production

Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

3 0

3 years ago

which of these solutions has the lowest freezing point? 0.25 m nacl 0.5 m nacl 1.0 m nacl 1.5 m nacl 2.0 m nacl

Alinara [238K]

The more particles (ions or molecules) that you can put into solution, the lower the freezing point.

the answer is E. 2.0 M nacl

4 0

3 years ago

Read 2 more answers

S + 6 HNO3 →→ H₂SO4 +6 NO₂ + 2 H₂O

bija089 [108]

Start with the 19.7 mol HNO3. use dimensional analysis to correctly convert from mol HNO3 to gram H2O. so, it should look similar to 19.7 mol HNO3 x (2 mol H2O/6 mol HNO3) x (18 g H2O/1 mol H2O)

the first parenthesis’ numbers were received from the balanced equation (for every 6 mol HNO3, 2 mol H2O formed). the second is converting from moles to grams by using the molar mass of H2O (1+1+16). you should get 709.2/6. once you divide those, the answer should be 118.2 g H2O. I’m not sure if your computer requires you to use the exact answer or stop at the correct number of significant digits, but if it does then it might just be 118. g H2O.

3 0

2 years ago

Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can

ratelena [41]

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = $Length\times breadth\times height$

Thus,

Volume = $304\times 0.014\times x\ mm^3=4.256x\ mm^3$

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

$Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3$

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

3 0

3 years ago