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3 years ago

Which is the frequency of the wave shown?

Chemistry

1 answer:

ratelena [41]3 years ago

5 0

Answer:

The frequency of wave is 0.125 Hz.

Explanation:

Frequency:

"It is an event repeat itself in a given period of time"

The unit of frequency is the Hz . If time is measured in seconds then frequency will be in Hz. Hz is equal to the per second.

Formula:

f = 1/ T

f = 1/ 8 sec

f = 0.125 Hz

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He predicted an element with an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum that he name

skad [1K]

Answer:

Scandium

Explanation:

Mendeleev played an important role in the development of the modern periodic table. His periodic table was filled with gaps. He said that these gaps were elements that were yet to be discovered. He rightly predicted many elements which have now been discovered and fitted in their proper places in the periodic table.

He used the prefix ''eka'' to refer to elements whose properties were alike but were yet to be discovered at that time.

The compound named ekaboron which he predicted to have an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum was later discovered in 1879 and properly named scandium.

3 0

3 years ago

"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92

Lynna [10]

Answer : The the concentration of $CaCl_2$ in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of $CaCl_2$ = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

$\fac{p^o-p_s}{p^o}=X_B$

where,

$p^o$ = vapor pressure of the pure component (water) = 92.6 mmHg

$p_s$ = vapor pressure of the solution = 83.1 mmHg

$X_B$ = mole fraction of solute, $(CaCl_2)$

Now put all the given values in this formula, we get the mole fraction of solute.

$\fac{92.6-83.1}{92.6}=X_B$

$X_B=0.102$

Now we have to calculate the mole fraction of solvent (water).

As we know that,

$X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898$

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, $(CaCl_2)$ and solvent, $(H_2O)$ .

$\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2$

$\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g$

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of $CaCl_2$

$Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%$

Therefore, the the concentration of $CaCl_2$ in mass percent is, 41.18 %

4 0

3 years ago

what mass of chromium (III) sulfide is produced from the reaction of 13.2 g of sulfur with excess chromium?

Andreas93 [3]

Answer: Cr2S3

Explanation:

5 0

3 years ago

Read 2 more answers

Is ink a compound or mixture?

Vladimir79 [104]

A mixture, because it contains multiple dyes and compounds

hopefully i could help ;)

8 0

3 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

3 years ago