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2 years ago

1. All non-living components of an ecosystem form a(n)

Chemistry

2 answers:

Salsk061 [2.6K]2 years ago

8 0

Explanation:

${ {.112201 \times \frac{?}{?} }^{?} }^{?} \times \frac{?}{?} \times \frac{?}{?}$

zubka84 [21]2 years ago

4 0

Abiotic environment

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The temperature is a 100

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3 years ago

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Relative formula mass of glucose? (C6H12O6)

Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

$C_{6}=6*(12.01g)=72.06g$

$H_{12}=12*(1.008g)=12.096g$

$O_{6}=6*(15.999g)=95.994g$

Then you add all of them together:

$72.06g+12.096g+95.994g=180.15g$

Therefore, the molar weight of glucose is 180.15 grams.

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3 years ago

What is the approximate mass of one proton?

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3 years ago

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P4O10 -> 4P+5O2 How many moles of phosphorus would be produced if 5.3 mol of P4O10 reacted?

viktelen [127]

Answer:

21.2 moles.

Explanation:

Hello!

In this case, for the given chemical reaction, we can see there is a 1:4 mole ratio between tetraphosphorous decaoxide and phosphorous; therefore, the following proportional factor provides the requested moles of phodphorous:

$n_P=5.3molP_4O_{10} *\frac{4molP}{1molP_4O_{10}} \\\\n_P=21.2molP$

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3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago