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kvv77 [185]
3 years ago
8

A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this

book in m3?
Chemistry
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

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RSB [31]

Answer : The number of molecules present in nitrogen gas are, 3.48\times 10^{13}

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of N_2 gas = 1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm      (1 atm = 760 mmHg)

V = Volume of N_2 gas = 985 mL = 0.982 L    (1 L = 1000 mL)

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 0.0^oC=273+0.0=273K

Now put all the given values in above equation, we get:

(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K

n=5.78\times 10^{-11}mol

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains 6.022\times 10^{23} number of molecules.

As, 1 mole of N_2 gas contains 6.022\times 10^{23} number of molecules

So, 5.78\times 10^{-11} mole of N_2 gas contains (5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13} number of molecules

Therefore, the number of molecules present in nitrogen gas are, 3.48\times 10^{13}

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3 years ago
Draw the lewis dot structures for each of the following molecules. show the resonance structures, if they exist
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A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled
dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$     \left(\frac{\text{mass}}{\text{volume}}\right)

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3 years ago
6. An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. To p
pishuonlain [190]

Answer:

[Na₂CO₃] = 0.094M

Explanation:

Based on the reaction:

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It is possible to find pH using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base,  [CO₃²⁻] = [Na₂CO₃] and  [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.

pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>

<em />

Replacing these values:

10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]

<em> [Na₂CO₃] = 0.094M</em>

<em />

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