What comes first in a thunderstorm?

the _ of an element is the average mass of an element naturally occurring atom, or isotopes,taking into account the _ of each is

NARA [144]

Answer:

the atomic mass of an element is the average mass of an element naturally occurring atom, or isotopes, taking into account the percentage of each isotopes

Explanation:

The atomic mass of an element is obtained by obtaining the relative abundances (in percentages) of naturally occurring atoms and the masses of the isotopes. The atomic mass can also be defined as the sum of the protons and the neutrons in the nucleus of an element.

In the periodic table, the atomic mass is indicated below the symbol of each of the elements and is usually in the decimal form.

7 0

2 years ago

If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef

andrezito [222]

Answer:

B

Explanation:

Recall the law of effusion:

$\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }$

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be r₁ and the rate of hydrogen be r₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for r₂:

$\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\ r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}$

Because there are 5 moles of hydrogen gas:

$\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}$

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0

2 years ago

Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

6 0

2 years ago

How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?

Assoli18 [71]

Answer:

$m_{Co^{3+}}=0.563gCo^{3+}$

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

$m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}$

Best regards!

4 0

2 years ago

When you burned copper, the product copper

MatroZZZ [7]

Answer:

Oxygen.

Explanation:

The copper must be combined with something in the air.

6 0

2 years ago

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