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1 year ago

Please help with these three questions.

1 answer:

3 0

B. 0.9 
D.Light intensity has no effect on whether electrons are emitted or not.
and
A. X=1.9eV,Y=0.2eV

I already took the gizmo so I know these are right. The first one I got wrong b/c there was no graph and the last one I didn't understand. Basically for the last one you calculate the work function for the metals and find their difference.

Hope this helps.

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consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q

Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado! :)

7 0

1 year ago

Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t

densk [106]

Answer:

Close to the calculated endpoint of a titration - Partially open

At the beginning of a titration - Completely open

Filling the buret with titrant - Completely closed

Conditioning the buret with the titrant - Completely closed

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

3 0

1 year ago

Analysis of the water content of a lake found in the desert showed that it contained 16.6 percent chloride ion, and had a densit

IRISSAK [1]

Answer : The molarity of the chloride ion in the water is, 5.75 M

Explanation :

As we are given that 16.6 % chloride ion that means 16.6 grams of chloride ion present 100 grams of solution.

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}$

$\text{Volume of solution}=\frac{100g}{1.23g/mL}=81.3mL$

Now we have to calculate the molarity of chloride ion.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of chloride ion}\times 1000}{\text{Molar mass of chloride ion}\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{16.6g\times 1000}{35.5g/mole\times 81.3mL}=5.75mole/L=5.75M$

Thus, the molarity of the chloride ion in the water is, 5.75 M

8 0

1 year ago

PLEASE HELP!!!!!!!!!!!!!

lions [1.4K]

Answer:

Pd(O2C2H3)2

Explanation:

Percentage composition of elements in the compound:

Pd= 47.40%

O=28.50%

C=21.40%

H= 2.69%

Molar mass of each element

Pd= 106g/mol

O= 16g/mol

C= 12g/mol

H= 1g/mol

Step 1

Find the moles of each element in 100g sample using the formula

no. of moles(n)=mass/molar mass

For Pd: 47.40g/106g/mol

n= 0.447 moles

For O: 28.50/16

n= 1.78 moles

For C: 21.40/12

n= 1.783 moles

For H: 2.69/1

n= 2.69moles

Step 2

Now divide the no. of moles separately by the smallest number of moles. Smallest number of moles is 0.447

Pd= 0.447/0.447

= 1

O= 1.78/0.447

= 3.9=4

C= 1.78/0.447

= 3.9=4

H= 2.69/0.447

= 6

Hence empirical formula is PdO4C4H6

The probable molecular formula is $Pd(O_{2}C_{2}H_{3})2$

8 0

1 year ago

Read 2 more answers

a.) The diameter of a uranium atom is 3.50 Å . Express the radius of a uranium atom in both meters and nanometers. b.) How many

Svetradugi [14.3K]

Answer:

Th answer to your question is:

a) 3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

1 A° ---------------- 1 x 10 ⁻¹⁰ m

3.5A° --------------- x

x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

1 A° ------------------ 0.1 nm

3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

6 0

2 years ago