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ladessa [460]
3 years ago
15

Oceń prawdziwość podanych informacji w układzie okresowym pierwiastki chemiczne są złożone według NIK to jących się liczbą atomo

wych Z Jądro atomowe jest bardzo mały i leży

Chemistry
1 answer:
cluponka [151]3 years ago
3 0
Wow. What language is this?
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Problem PageQuestion A major component of gasoline is octane . When octane is burned in air, it chemically reacts with oxygen ga
Ivahew [28]

Answer:

grams of oxygen = 21.40 g

Explanation:

octane → C8H18

The chemical reaction needs to be represented with a chemical equation before solving. The equation also need to be balanced.

C8H18 + O2 → CO2 + H2O

Balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Molar mass of octane = 12 × 8 + 18 = 96 + 18 = 114 g

Mass of octane in the chemical equation = 2 (114) = 228 g

molar mass of oxygen = 32 g

mass of oxygen in the chemical equation = 25 × 32 = 800  g

if  228 g of octane react with 800 g of oxygen

6.1 g of octane will require ? grams of oxygen

cross multiply

grams of oxygen = 6.1 × 800/228

grams of oxygen = 4880 /228

grams of oxygen = 21.4035087719

grams of oxygen = 21.40 g

 

5 0
3 years ago
The molar mass of a certain gas is 49 g. What is the density of the gas in g/L at STP?
snow_tiger [21]

Answer:

\boxed{\text{2.2 g/L}}

Explanation:

We can use the Ideal Gas Law to calculate the density of the gas.

   pV = nRT

      n = m/M           Substitute for n

   pV = (m/M)RT     Multiply both sides by M

pVM = mRT            Divide both sides by V

  pM = (m/V) RT

     ρ = m/V             Substitute for m/V

 pM = ρRT              Divide each side by RT

\rho = \frac{pM }{RT}

Data:

p = 1.00 bar

M = 49 g/mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 0 °C = 273.15 K

Calculation:

ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L

The density of the gas is \boxed{\text{2.2 g/L}}.

8 0
4 years ago
What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor
Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

CH_3COOH(l)\rightarrow CH_3COOH(g)..[1]

Heat of vaporization of acetic acid = H^o_{vap}=24.3 kJ/mol

CH_3COOH(s)\rightarrow CH_3COOH(l)..[2]

Heat of fusion of acetic acid = H^o_{fus}=10.8 kJ/mol

Heat of sublimation of acetic acid = H^o_{sub}=?

CH_3COOH(s)\rightarrow CH_3COOH(g)..[3]

[1] + [2] = [3] (Hess's law)

H^o_{sub}=H^o_{vap}+H^o_{fus}

=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0
3 years ago
Worth 40 Points❤ 10th grade chemistry ‍
brilliants [131]

Answer:

  1. F
  2. E
  3. G
  4. A
  5. C
  6. B
  7. D

Explanation:

Those are the answers in order, BUT the Goldstein and Rutherford ones are confusing me because Rutherford discovered<em> protons</em> and the <em>nuclear atom</em>. Through my research I also noticed that Goldstein contributed to the discovery of the protons made Rutherford, so I listed him as the one who discovered the proton.

I hope this helps! Have a wonderful night! :D

6 0
3 years ago
Which of the following pairs of molecules can be held together by a hydrogen bond??
OverLord2011 [107]
<span>CH3OH and NH3
I HOPE THIS HELPS :D</span>
4 0
3 years ago
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