All categories

3 years ago

Calculate the mass (in g) of 2.1 x 1024 atoms of W.

1 answer:

6 0

Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams

You might be interested in

Temperature has no significance during the calculation of numerical .why?? <br>^_^

Brums [2.3K]

Answer:

Explanation:

concepts, such as the internal energy of a system; heat or sensible heat, which are defined as types of energy transfer (as is work); or for the characteristic energy of a degree of freedom in a thermal system {\displaystyle kT}kT, where {\displaystyle T}T is temperature and {\displaystyle k}k is the Boltzmann constant.

7 0

3 years ago

Where are the transition metals on the periodic table?

densk [106]

D. In columns 3-12 in the centre of the table

3 0

3 years ago

A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse

ludmilkaskok [199]

Answer:

$\large \boxed{\text{5.9 s}}$

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of effusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

$\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}$

4 0

3 years ago

Which of these factors should remain constant in a simulation of natural

musickatia [10]

A is your answer The ratio of females to males

8 0

1 year ago

How does lacquer thinner work chemically?

Anton [14]

Have a nice day today

8 0

3 years ago