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Dennis_Churaev [7]
3 years ago
8

Directions: Perform the Assessment in Module 7 because that will serve as your Performance Task for Module 5, 6 & 7. Kindly

read the rubric for you to be guided on how your output will be graded. Do not forget to submit your output to your teacher. Enjoy!
please answer nyo to nang maayos please ​

Chemistry
1 answer:
garik1379 [7]3 years ago
8 0

please put your question completely....

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how many atoms of phosphorous are in 4.90 mol of copper(ii) phosphate? the formula for copper(ii) phosphate is Cu3(PO4)2.
Karolina [17]

Answer:

\boxed{5.90 \times 10^{24}}

Explanation:

Step 1. Calculate the formula units of Cu₃(PO₄)₂  

\text{4.90 mol Cu$_{3}$(PO$_{4}$)$_{2}$} \times \dfrac{6.022 \times 10^{23} \text{ formula units}}{\text{1 mol Cu$_{3}$(PO$_{4}$)$_{2}$}}\\\\=\text{2.951 $\times$ 10$^{24}$ formula units Cu$_{3}$(PO$_{4}$)$_{2}$}

Step 2. Calculate the atoms of P

\text{Atoms of P}\\\\=\text{2.951 $\times$ 10$^{24}$ formula units Cu$_{3}$(PO$_{4}$)$_{2}$} \times \dfrac{ \text{2 atoms P}}{\text{1 formula unit Cu$_{3}$(PO$_{4}$)$_{2}$}}\\\\= \boxed{5.90 \times 10^{24} \text{ atoms P}}

8 0
3 years ago
Hi, pls can someone answer<br> I'll give brainliest
Temka [501]

D is the answer so now we can understand

6 0
2 years ago
Read 2 more answers
A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
6 0
2 years ago
Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate t
LenaWriter [7]
The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He. 
To fill a volume of 6.3 L the mass of He required 
    = 1.79 x 10⁻⁴ g/mL * 6300 mL
   = 11 277 * 10⁻⁴ g
  Therefore mass of He required = 1.1277 g of He
8 0
3 years ago
Read 2 more answers
Upon arrival we needed to hunt in this new land we only had five refills and they needed 50 g of gunpowder to be shot once. We o
emmasim [6.3K]

Answer:

Explanation:

Upon arrival we needed to hunt in this new land we only had five refills and they needed 50 g of gunpowder to be shot once. We only have 15 pounds of gunpowder. It is taking six shots to kill one of these wild turkeys. How many turkeys can be shot with 15 pounds of gunpowder?

If we had plenty of refills, and it takes 6 shots to kill a wild turkey at 50 gms of gunpowder per shot, then each turkey requires 6X50 =300gms of gunpowder.  We have 15X454 gms of gunpowder and have the potential to kill 15X454/300=22.7 or 22 turkeys.and it takes 6 shots to kill a wild turkey.

The limiting reagent is the number of refills, and withonly 5, we are out of luck and can't kill one turkey

6 0
2 years ago
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