Heeeeellllllppppppp!!!!!

I need help pls its due tomorrow i wanna get the answer right pls.

marin [14]

Answer:

1) Without our atmosphere we wouldn't have life! The atmosphere protects our earth from letting harmful substances or rocks into our planet, which could cause damage.

2) You interact with the atmosphere in both good ways and bad. You interact by putting up things like flowers and trees, but in bad ways by releasing bad things (like carbon dioxide) and by light pollution!

3) The factors that can affect climate of an area, by latitude and longitude!

Explanation:

Hope this helps! Plz mark as brainliest! :)

6 0

3 years ago

Question 1 (1 point)

andrezito [222]

Answer:

True 
B. 24 hours 
Light and Dark 
A. To spin on an Axis 
True 
Nearest P0INTS 
B. 365 days 
True 
A. January 
B. July 
Summer 
B. The tilt of the Earth on its axis
True 
¿-¿
¿-¿
B. June
A. September 
B. December and March

Explanation:

Sorry yan lang alm but I hope it helps

<h2>#CARRYONLEARNING </h2><h2>#STUDYWELL </h2>

5 0

3 years ago

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10 + O2----------------> CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A

8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm

6 0

3 years ago

PLEASE HELPPPPPPPPPPPPPP

melamori03 [73]

It changes from a liquid from to a solid.

I hope this helps!

3 0

3 years ago

A compound is composed of only C, H, and O. The combustion of a .519-g sample of the compound yields 1.24 g of CO_2 and .255 g o

Kay [80]

Answer:

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of compound = 0.519 g

Mass of CO₂ = 1.24 g

Mass of H₂O = 0.255 g

Empirical formula = ?

Solution:

%age of C,H,O

C = 1.24 g/0.519 × 12/44 ×100 = 65.5%

H = 0.255 g/0.519 × 2.016/18 ×100 = 5.6%

O = 100 - (65.5+5.6)

O = 28.9%

Number of gram atoms of H = 5.6 / 1.01 = 5.5

Number of gram atoms of O = 28.9 / 16 = 1.81

Number of gram atoms of C = 65.5 / 12 = 5.5

Atomic ratio:

C : H : O

5.5/1.81 : 5.5/1.81 : 1.81/1.81

3 : 3 : 1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

8 0

4 years ago