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Crank
3 years ago
10

Plant use sunlight as energy to convert carbon dioxide and water into glucose and oxygen. Which best describes the reaction?

Chemistry
2 answers:
Illusion [34]3 years ago
8 0

Answer:

Its the process of photosynthesis

motikmotik3 years ago
3 0

Answer:

photosynthesis(definitely the answer) or photolysis(different, slightly, but kinda the same thing)

Explanation:

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Colorful flowers and blueberrie looking ones and the trees
4 0
3 years ago
Read 2 more answers
4/5(2x 5)−4=1
uysha [10]
 5/8=0.625 4/5=.8


5/8 because 5/8=0.625 and because its 2x you multiply 0.625 by 2=1.25 then you do 1.25+5=6.25 4/5=.8    6.25 times .8=5 5-4=1   


X=5/8 or 0.625

5 0
3 years ago
A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1°C for four-phase equilibrium of allotropic solid
Inessa [10]

Answer:

The claim is untrue.  

Explanation:

Given the information from the question. We need to evaluate the claim.

According to the phase rule we have F= C-P+2

In this particular situation, the forms are completely allotropic .In order words, they are conjured through the same chemical composition. Thus constituents C= 1, P=4 for four phases and the number variables is 2. As a result, F= C-P+2 =1-4+2= -1. Therefore, the claim is untrue.  

8 0
3 years ago
A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0
3 years ago
[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con
telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = 284 \ grams

solid soil volume =205 \ cc

saturated mass soil = 361 \ g

The weight of the soil after drainage is =295 \ g

Water weight for soil saturation = (361-284) = 77 \ g

Water volume required for soil saturation =\frac{77}{1} = 77 \ cc

Sample volume of water: = \frac{\text{water density}}{\text{water density input}}

= 361- 295 \\\\ = 66 \ cc

Soil water retained volume = (draining field weight - dry soil weight)

                                             = 295 - 284 \\\\ = 11 \ cc.

\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}

                    = \frac{77}{(205 + 77)}  \\\\= \frac{77}{(282)}  \\\\ = 27.30 \%

(Its saturated water volume is equal to the volume of voids)

\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}

                              = \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23

\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}

                            = \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04

6 0
3 years ago
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