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3 years ago

If a wave is traveling at a steady speed and its frequency is

Chemistry

1 answer:

Nataly [62]3 years ago

4 0

Answer:

If the frequency is doubled, the wavelength is only half as long.

Explanation:

Wave speed equals frequency*wavelength. So doubling the frequency must halve the wavelength in order for wave speed to remain the same.

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"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

4 0

3 years ago

An atom of calcium contains 20 protons how many electrons does it have

harina [27]

20.

Atomic number is equivalent to its protons and electrons. :)

4 0

3 years ago

ICl has a higher boiling point than Br2. What is the best explanation for this?

Flura [38]

<span>The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.</span>

7 0

3 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T $_2$ ),

P $_1$ V $_1$ / T $_1$ = P $_2$ V $_2$ / T $_2$ ,

T $_2$ = P $_2$ V $_2$ T $_1$ / P $_1$ V $_1$ ,

T $_2$ = 0.73 atm $*$ 1.8 L $*$ 298.15 K / 1 atm $*$ 1.2 L = ( 0.73 $*$ 1.8 $*$ 298.15 / 1 $*$ 1.2 ) K = 326.47425 K,

T $_2$ = 326.47425 K = 53.32425 C

7 0

3 years ago

Concentrated HCl is 37% m/m. The density of the solution is 1.19 g/mL. Assume you have 100.0 grams of SOLUTION.

sdas [7]

Answer:

M HCl sln = 12.0785 M

Explanation:

molarity (M) [=] mol/L
%mm = ((mass compound)/(mass sln))*100

∴ mass sln = 100.0 g

∴ δ sln = 1.19 g/mL

∴ % m/m = 37 %

⇒ 37 % =((mass HCl/mass sln))*100

⇒ 0.37 = mass HCl / 100.0 g

⇒ 37 g = mass HCl

∴ molar mass HCl = 36.46 g/mol

⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol

⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L

⇒ M HClsln = 1.015 mol/0.084034 L

⇒ M HCl sln = 12.0785 M

7 0

3 years ago