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3 years ago

Calculate the period when the frequency is 50Hz.

Chemistry

1 answer:

liraira [26]3 years ago

8 0

Answer:

For example, a wave with a time period of 2 seconds has a frequency of 1 ÷ 2 = 0.5 Hz. A radio wave has a time period of 0.0000003333333 seconds.

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Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you ho

german

Answer:

The value is $Z = 311.33 \ g/mol$

Explanation:

From the question we are told that

The mass of saxitoxin is $m = 12.5 mg = 12.5 * 10^{-6} g$

The volume of water is $V = 3.10 mL = 3.10 *10^{-3} L$

The osmotic pressure is $P = 0.236 = \frac{0.236}{760} = 3.105 * 10^{-4} atm$

The temperature is $T = 19^oC = 19 + 273 = 292 \ K$

Generally the osmotic pressure is mathematically represented as

$P = C * T * R$

Here R is the gas constant with value

$R = 0.0821 ( L .atm /mol. K)$

and C is the concentration of saxitoxin

$3.105 * 10^{-4} = C * 0.0821 * 292$

$C = 1.295 *10^{-5} mol/L$

Generally the number of moles of saxitoxin is mathematically represented as

$n = C * V$

=> $n = 1.295 *10^{-5} *3.10 *10^{-3}$

=> $n = 4.015 *10^{-8} \ mol$

Generally the molar mass of saxitoxin is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}$

=> $Z = 311.33 \ g/mol$

5 0

4 years ago

Each liter of air has a mass of 1.80 grams. How many liters of air are contained in

natita [175]

Question:

Each liter of air has a mass of 1.80 grams. How many liters of air are contained in 2.5 x103) kg of air?

Answer:

11.48106 L

5 0

3 years ago

What is the difference between a permanent magnet and a temporary magnet

vichka [17]

If something becomes magnetized for a short amount of time it is only a matter of time before it isn't magnetic anymore. permeant on the other hand is forever and will never undo itself.

3 0

3 years ago

Calculate ΔGrxn for this equation, rounding your answer to the nearest whole number. CaCO3(s) → CaO(s) + CO2(g) ΔGf,CaCO3 = −1,1

FrozenT [24]

Answer: $\Delta G_{rxn}$ for the given reaction is 130.19kJ/mol

Explanation:

To Calculate the $\Delta G_{rxn}$ , we use the formula:

$\Delta G_{rxn}=\Delta G_{products}-\Delta G_{reactants}$

For the given chemical reaction:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

We are given:

$\Delta G_{CaCO_3}=-1128.76kJ/mol\\\Delta G_{CaO}=-604.17kJ/mol\\\Delta G_{CO_2}=-394.4kJ/mol$

Now, to calculate $\Delta G_{rxn}$ , we put the values in the above equation:

$\Delta G_{rxn}=(\Delta G_{CaO}+\Delta G_{CO_2})-\Delta G_{CaCO_3}$

$\Delta G_{reaction}=(-604.17-394.4)-(-1128.76)kJ/mol\\\Delta G_{reaction}=130.19kJ/mol$

As, the value of $\Delta G_{reaction}$ comes out to be positive, the reaction is said to be non-spontaneous reaction.

3 0

3 years ago

6) What types of atomic orbitals are in the third principal energy

Bumek [7]

Answer:

Types of atomic orbitals present in the third principal energy are s, p and d only .

Explanation:

OPTION A-: s and p atomic orbitals - these two orbitals are present in second principal energy level. Therefore , the option is incorrect.

 OPTION B-: p and d only - This option is wrong as there is no such principal level energy where , s atomic orbital is absent .

OPTION C-: s , p and d only -these orbitals are present in third principal energy level. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .

OPTION D-: s , p, d and f only -There is also a f sublevel at the fourth and higher stages, containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .

Thus , the correct option is C (s , p and d only .)

6 0

4 years ago

Calculate the period when the frequency is 50Hz.​

Calculate the period when the frequency is 50Hz.