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aksik [14]
3 years ago
12

What are three things that living things need to live

Chemistry
2 answers:
Grace [21]3 years ago
8 0
Food shelter oxygen
nasty-shy [4]3 years ago
3 0

water, food, and air.

Additionally:

Shelter and clothes

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What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?
tensa zangetsu [6.8K]

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

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3 0
3 years ago
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Which of the following is a reasonable ground-state electron configuration?
sladkih [1.3K]
The reasonable ground-state electron configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 4d8
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2 years ago
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Which sentences describe decomposers in a food chain?
Elan Coil [88]

Answer:

hey are the final link in the energy flow in a food chain or a food web. They are fungi and animals that feed on dead organic mattet.

Explanaticuz im naruto uzamaki

5 0
2 years ago
The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5
vazorg [7]

A) c = 3 x 10^8 m/s 
f = 7.15 x 10^14 Hz 
c = λ x f (=) λ = 3 x 10^8 / 7.15 x 10^14 = 4.19 x 10^-7 m = 419.6 nm 

B) E = h  f 
H = Planck's constant = 6.63 x 10^-34 J/s 
E = 6.63 x 10^-34 x 7.15 x 10^14 = 4.74 x 10^-19 J

Read more on Brainly.com - brainly.com/question/5760368#readmore
3 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
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