All categories

3 years ago

Which element can join with other elements to form covalent bonds? beryllium (Be) carbon (C) cesium (Cs) iron (Fe)

Chemistry

2 answers:

Alex_Xolod [135]3 years ago

5 0

Answer:

Carbon (C)

Explanation:

Carbon is the only nonmetal there and covalent bonds happen between two nonmetal atoms

sineoko [7]3 years ago

3 0

Answer:

It’s C

Explanation:

You might be interested in

A truck tire has a volume of 218 L and is filled with air to 35.0 psi at 295 K. After a drive, the air heats up to 318 K. (b) If

Alenkasestr [34]

The pressure (in psi) is 37.1

We are given the following information:

- The initial temperature of the air, $T_{1}=295 \mathrm{~K}$

- The initial pressure, $P_{1}=35.0 \mathrm{psi}$ .

- The initial volume, $V_{1}=208 \mathrm{~L}$

- The final temperature, $T_{2}=319 \mathrm{~K}$

- The increase in the volume is $2 \%$ , that is, $\Delta V=\frac{2}{100} V_{1}$ where $\Delta V$ is the increase in the volume.

The combined gas law states that a fixed amount of an ideal gas obeys the following equation: $\frac{P V}{T}=$ constant, where:

- P is the Pressure of the gas.

- V is the Volume of the gas.

- n is the number of moles of gas.

$R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}=0.0821 \mathrm{~L} \cdot {atm} / \mathrm{mol} \cdot \mathrm{K}$ is the Universal Gas constant.

- T is the absolute temperature of the gas.

The final volume of the air is:

$\begin{aligned}V_{2} &=V_{1}+\Delta V \\&=V_{1}+\frac{2}{100} V_{1} \\&=\frac{102}{100} V_{1}\end{aligned}$

Equating the initial and final state, we have:

$\begin{aligned}\frac{P_{2} V_{2}}{T_{2}} &=\frac{P_{1} V_{1}}{T_{1}} \\\Rightarrow P_{2} &=\frac{V_{1}}{V_{2}} \times \frac{T_{2}}{T_{1}} \times P_{1} \\&=\frac{V_{1}}{102 V_{1} / 100} \times \frac{319 \mathrm{~K}}{295 \mathrm{~K}} \times 35.0 \mathrm{psi} \\& \approx 37.1 \mathrm{psi}\end{aligned}$

The ideal gas law states that a universal constant for an ideal gas is the ratio of the product of pressure and temperature to the product of the number of moles and absolute temperature. The resultant equation is known as the combined gas law if the number of moles in the ideal gas law is set to a constant.

Learn more about combined gas law brainly.com/question/13154969

#SPJ4

8 0

2 years ago

Eventually, everyone must sleep due to our (1 point) need for mental rest. beta waves. biological rhythms. delta waves.

Bad White [126]

Out of the available answers, the most likely correct would be that eventually everyone must sleep due to our need for mental rest. However, it is not as simple as this, as all of the processes implicated with mental rest (including the restoration of biological rhythms as mentioned in an alternative answer) cannot necessarily be summed up in this way.

6 0

3 years ago

Is chex mix an example of a pure substance or mixture?

irinina [24]

The answer is mixture

3 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

i need to kono why this app is so ....i get a add then i hav ea nohter add and there way to long. anyways 3 more weekds till sum

erastova [34]

Answer:

same I hate all the frickin ads ;-;

you gave a good day too mate

3 0

3 years ago

Read 2 more answers