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Nutka1998 [239]
2 years ago
7

a classroom has an aquarium. In the aquarium are a goldfish and a water plant. the teacher explains that both organisms use cell

ular respiration to transform food into energy. identify how the plant and the goldfish get the materials needed for cellular respiration.
Chemistry
1 answer:
Anna71 [15]2 years ago
4 0
the thing that helps is always gonna help that’s why u should always help
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frez [133]

The smallest functional and structural unit of an organism, usually microscopic and consisting of cytoplasm and a nucleus in a membrane.

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WHEN I SAY I NEED HELP ASAP !!!
sweet [91]

Answer:

monoxide

dioxide

trioxide

tetroxide

pentoxide

hexoxide

heptoxide

octoxide

nonoxide

Explanation:

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3 years ago
002 (part 1 of 2) 10.0 points
BARSIC [14]

Answer:

4) 1.5 mol

Explanation:

Well, the equation is already balanced and the mole to mole ratio of reactants and products are all 1. So if the limiting reactant is HCl and you have 1.5 mol, you do the mole to mole ratio with NaCl and since it is 1 to 1, there'd be 1.5 mol of NaCl.

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3 years ago
Assuming all volume measurements are made at the same temperature and pressure, how many liters of oxygen gas would it take to r
EleoNora [17]
2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625
3 0
3 years ago
Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac
stich3 [128]

Answer:

Value of Q_{p} for the given redox reaction is 1.0\times 10^{-8}

Explanation:

Redox reaction with states of species:

14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)

Reaction quotient for this redox reaction:

Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of H_{2}O is taken as 1 due to the fact that H_{2}O is a pure liquid.

pH=-log[H^{+}]

So, [H^{+}]=10^{-pH}

Plug in all the given values in the equation of Q_{p}:

Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}

7 0
3 years ago
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