All categories

3 years ago

Two isotopes of the same element must have the same

Chemistry

1 answer:

emmainna [20.7K]3 years ago

3 0

Answer:

atomic number

Explanation:

isotope is a family of an element with the same atomic number but different mass number.

You might be interested in

3. Given the following equation:

miskamm [114]

Answer:

4.767 grams of KCl are produced from 2.50 g of K and excess Cl2

Explanation:

The balanced equation is

2 K+ Cl2 --->2 KCI

Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K

Mass of one atom/mole of potassium is 39.098 grams

Number of moles is 2.5 grams = $\frac{2.5}{39.098} = 0.064$

So, 2 moles of K produces 2 moles of KCL

0.064 moles of K will produces 0.064 moles of KCl

Mass of one molecule of KCl is 74.5513 g/mol

Mass of 0.064 moles of KCl is 4.767 grams

4 0

3 years ago

The density of a metal is 9.80 g/mL. What is the mass of a sample of metal when dropped in 28.9 mL of water, the volume increase

yan [13]

The correct answer is 2.70 × 10² g or 270 g.

It is given, that the density of a metal is 9.80 g/ml.

Let the mass of a sample of metal be x.

The sample of metal is dropped in 28.9 ml of water, due to which the volume of the water increases to 56.4 ml.

In order to calculate the mass of a metal, there is a need to use the formula, mass = density * volume

Mass = (9.80 g/ml) (56.4 ml - 28.9 ml)

= (9.80 g/ml) (27.5 ml)

= 2.70 × 10² g or 270 g

6 0

3 years ago

Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K.

Montano1993 [528]

Explanation:

Relation between pressure of water and its droplet is as follows.

$ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}$

where, p = pressure of droplet

$p_{o}$ = water pressure in given temperature

$\gamma$ = $7.99 \times 10^{-3}$

M = Molecular Weight in Kg/Mol (0.018 for water)

r = radius in meters

$\rho$ = density of water in $Kg/m^{3}$ (1000 $kg/m^{3}$ )

R = ideal gas constant (8.31)

T = temperature in Kelvin

(a) We will calculate the value of p as follows.

p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

= $e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2$

= 26.8 torr

(b) And, vapor pressure of spherical water droplets of radius 2.0 $\mu m$ or $2 \times 10^{-6} m$

p = $e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}$

= $e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2$

= 25.2 torr

7 0

4 years ago

What would be the effect on the calculated ratio (mole of copper:mole of iron) if:

kotegsom [21]

Explanation:

Since the ratio is Cu/Fe, if some Fe were lost due to spillage, the Cu/Fe ratio would INCREASE because Fe would be lower.

5 0

3 years ago

How does the energy of an electron change when the electron moves closer to the nucleus?

makkiz [27]

The energy of the electron reduces as it moves closers to the nucleus. This is because it moves from a higher energy orbital to a one of lower energy. It does this by releasing a photon with the energy difference of the current energy level, and the energy level it drops too.

6 0

3 years ago

Read 2 more answers