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iVinArrow [24]
2 years ago
12

How long does it take the Earth to move from Point A to Point C?

Chemistry
2 answers:
e-lub [12.9K]2 years ago
8 0

Answer:

24 hours

Explanation:

ruslelena [56]2 years ago
5 0

Answer:182 days and 5 hours

Explanation: it takes 365 days for the earth to go around the sun so 365 divided by 2 = 182 days and 5 hours.

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Each of the following organs had a role in excretion, EXCEPT for the _____?
eimsori [14]

Answer:

c. the heart

all the rest actually will excrete, the heart is in a closed system so contributes to excretory organs but doesn't actually do it barring an accident

Explanation:

8 0
2 years ago
Read 2 more answers
Human blood contains a buffer of carbonic acid (H_2 CO_3) and bicarbonate anion (HC〖O_3〗^-) in order to maintain blood pH betwee
kirill115 [55]

Kc = [H3O+][HCO3-] / [H2CO3]

Remember that Kc is products over reactants. Also, you do not include liquid water in a Kc expression, since liquid water has no concentration.

6 0
2 years ago
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
2 years ago
PLEASE ANSWER ASAP <br> please get it right
musickatia [10]

Answer: C

Earths atmosphere maintains a stable temp

4 0
3 years ago
What is the empirical formula for a compound that contains 38.77% Cl and 61.23% O?
Leno4ka [110]
W(Cl)=0.3877
w(O)=0.6123
M(Cl)=35.5 g/mol
M(O)=16.0 g/mol

ClₐOₓ
M(ClₐOₓ)=35.5a+16.0x

w(Cl)=35.5a/(35.5a+16.0x)
w(O)=16.0x/(35.5a+16.0x)

solve a system of two equations with two unknowns
35.5a/(35.5a+16.0x)=0.3877
16.0x/(35.5a+16.0x)=0.6123

a=2
x=7

Cl₂O₇ is the empirical formula
3 0
3 years ago
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