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3 years ago

15

If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane mirror, how far from the mirror is the

image located? Explain your reasoning.

1 answer:

ryzh [129]3 years ago

7 0

Answer:

the mirror is 12 cm away from the image

Explanation:

; the image will be virtual and it will form at 12 cm distance behind the mirror.

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Water can only dissolve inorganic compounds

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Calculate the mass of CaCl2•2H2O required to make 100.0 mL of a 0.100 M solution. Each of the calculations below will take you t

Zolol [24]

Answer:

The mass is 1.4701 grams and the moles is 0.01.

Explanation:

Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,

Molarity = moles of solute/volume of solution in liters

Now putting the values we get,

0.100 = moles of solute/0.1000

Moles of solute = 0.100 * 0.1000

= 0.01 moles

The mass of CaCl2.2H2O can be determined by using the formula,

Moles = mass/molar mass

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3 years ago

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et

Alexxx [7]

<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For testosterone:</u>

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol$

<u>For diethyl ether:</u>

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}$

$\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095$

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

$p^s$ = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

$\chi_{\text{solute}}$ = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

$\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg$

Hence, the vapor pressure of solution is 459.17 mmHg

7 0

3 years ago