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Mkey [24]
3 years ago
6

Help please really need help

Chemistry
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

2.14 × 10⁻³ molecules/RSP

3.31 × 10⁻³ molecules/ESP

Explanation:

Step 1: Calculate the number of moles of Acetaminophen per Regular Strength Pill (RSP)

A Regular Strength Pill has 1.29 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in  1 mole of molecules.

1.29 × 10²¹ molecules/RSP × 1 mol/6.02 × 10²³ molecules = 2.14 × 10⁻³ molecules/RSP

Step 2: Calculate the number of moles of Acetaminophen per Extra Strength Pill (ESP)

An Extra Strength Pill has 1.99 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in  1 mole of molecules.

1.99 × 10²¹ molecules/ESP × 1 mol/6.02 × 10²³ molecules = 3.31 × 10⁻³ molecules/ESP

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An element has ccp packing with a face-centered cubic unit cell. its density is 8920 kg/m3 and the unit cell volume is 4.72 x 10
ziro4ka [17]
<span>63.4 g/mol
   First, let's determine how many atoms per unit cell in face-centered cubic. There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms. Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
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3 years ago
Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---&gt; C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation
nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ...[1]

C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ..[2]

H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ..[3]

2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}

\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)

\Delta H^o_{4}=232 kJ

The enthalpy for given reaction is 232 kilo Joules.

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Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t
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Answer:

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Explanation:

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I hope you undestand me

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