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3 years ago

14

Density is _____ to pressure.

2 answers:

In-s [12.5K]3 years ago

8 0

C. not related
density is lower.

irinina [24]3 years ago

6 0

The answer is A. Directly Proportional :)

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How many total atoms are in 0.290 g of P2O5?

Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

$\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\$

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

$\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\$

$= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\$

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ $\text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\$

8 0

3 years ago

sashaice [31]

I think it's DNA forms a double helix and RNA consists a single strand.
I may be wrong...

3 0

2 years ago

Read 2 more answers

Copper2 nitrate > nitrogen dioxide + copper2 oxide+ oxygen

erastovalidia [21]

Cu(NO3)2>NO2+CuO+O2 balanced: 2Cu(NO3)2=4NO2+2CuO+O2

3 0

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]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

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

Which would most likely cause a person to produce antibodies?

rosijanka [135]

Answer:

receiving a vaccination

Explanation:

7 0

3 years ago

Read 2 more answers