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Ierofanga [76]
3 years ago
13

39. Period is the force of attraction between objects resulting in objects pulling towards each

Chemistry
1 answer:
storchak [24]3 years ago
8 0

Answer:

True

Explanation:

if there was nothing to avoid of objects pulling towards one another many results may happen

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What is the pH of a solution with a concentration of 2.4 × 10-5 molar H3O+? Show, or explain, the work used to solve this proble
MrRissso [65]
pH=-log|H_{3}O^{+}|\\\\
pH=-log2.4*10^{-5}=-(log2.4+log10^{-5})=-(0,38-5)=-(-4.62)\\\\
pH=4.62
7 0
3 years ago
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In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5
Rufina [12.5K]

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

7 0
2 years ago
Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? a. Mg b. Li c. Al d. Pb
Stells [14]

I believe the answer is A) Mg.

5 0
3 years ago
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Screen of monitors and televisions are made by ?
kati45 [8]

Answer:

C

Explanation:

This is a correct answer i think soo

8 0
2 years ago
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the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
shtirl [24]

<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

4 0
3 years ago
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