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2 years ago

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Chemistry

1 answer:

NemiM [27]2 years ago

3 0

Answer:

7 false

8 True

Explanation:

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Question 4 (1 point)<br> If you have exactly one million atoms how many moles do you have?

rjkz [21]

Answer:

1.66 × 10⁻¹⁸ Moles

Explanation:

As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

Number of Moles = 1.66 × 10⁻¹⁸ Moles

7 0

3 years ago

Which medical condition(s) can prevent molecules from entering the cell?

kvasek [131]

Answer:

Explanation:

any type of spreading disease that kills

6 0

2 years ago

Read 2 more answers

What is the entropy change of the surroundings

KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of acetone = 58 g/mol

Number of moles = $\frac{10.8}{58}=0.1862moles$

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = $\frac{32.3}{1}\times 0.1862=5.828kJ/mol$

To calculate the entropy change for the system, we use the formula:

$\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}$

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

$\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol$ (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

$\Delta S_{system}+\Delta S_{surrounding}=0$

$\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol$

5 0

3 years ago

A nuclear fission reaction has mass difference between the products and the reactants of 0.0284 amu. Calculate the

notsponge [240]

Answer:

C) 4.24 x $10^{-12}$

Explanation:

E = Δm $c^{2}$

Δm = 0.0284 x 1.66 x $10^{-27}$ kg = 4.714x $10^{-29}$ kg

putting value in above equation

E = 4.714x $10^{-29}$ kg x (3x $10^{8}$ $)^{2}$ = 4.24 x $10^{-12}$

6 0

2 years ago

Calculate the mass percent of oxygen in CO2<br> (Work shown)

Genrish500 [490]

Answer:

$mass \: of \: CO _{2} = (12 + (16 \times 2)) = 44 \: g \\ mass \: of \: oxygen \: in \: CO _{2} = (16 \times 2) = 32 \\ \% \: of \: oxygen = \frac{32}{44} \times 100\% \\ = 72.7\%$

4 0

2 years ago