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zhuklara [117]
3 years ago
15

CO2(g) + H2O(I) - -> C6H12O6(s) + O2(g)

Chemistry
1 answer:
Lana71 [14]3 years ago
4 0

Answer:

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Explanation:

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Answer:

the answer is C

Explanation:

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What are three things that living things need to live
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QUICK HELP PLEASE!!?? Need answer QUICK
d1i1m1o1n [39]

Answer:

B

Explanation:

Look on the x-axis for the tick marked "60". This indicates 60 degrees Celsius, which we want. Now, look on the y-axis for the tick marked "60". This indicates 60 grams of Na_2HAsO_4. Trace along the graph to find where these two places meet at (60, 60).

Now, look for the solubility curve of Na_2HAsO_4; it's the yellow-orange line. Find out what the y-coordinate of the point where x = 60 is on the line: it's around (60, 65).

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If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
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So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
4 0
3 years ago
Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were intr
dalvyx [7]

Explanation:

The given data is as follows.

      V_{1} = 250 mL,     V_{2} = 750 mL

      T_{1} = 35^{o}C = 35 + 273 K = 308 K

      T_{2} = 35 + 273 K = 308 K

      P_{1} = 0.55 atm,    P_{2} = 1.5 atm

               P = ? ,         V = 10.0 L

Since, temperature is constant.

So,    P_{1}V_{1} + P_{2}V_{2} = PV

Now, putting the given values into the above formula as follows.

         P_{1}V_{1} + P_{2}V_{2} = PV

         0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm = P \times 10.0 L

                     P = 0.126 atm

As, 1 atm = 760 torr. So, 0.126 atm \times \frac{760 torr}{1 atm} = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.

8 0
3 years ago
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