List two major properties of most metals on the periodic table

Avogadro's law states that the _____ of a gas is directly proportional to the _____ of the gas when _____ stay the same

attashe74 [19]

Answer:

The three blanks for this answer, are

1. volumen

2. moles

3. Temperature and pressure.

So, Avogadro's law states that the volume of a gas is directly proportional to the moles of the gas when temperature and pressure stay the same

Explanation:

Imagine you have 10 moles of a gas which is contained in 50 L. How many moles of that gas, you will have if the volumen has been reduced to 10 L. (Of course, don't forget that T° and pressure are the same)

There is an equation like this, initial moles /initial volume = moles at the end/volume at the end, (Avogadro law for gases), so 10/50 =moles at the end/10. When u operate, moles at the end = (10 x 10) / 50.

Moles at the end are 2. Did u get it?. Volumen has been reduced, also the moles.

7 0

3 years ago

10 POINTS : What are 4 things created from Big Bang?

Leno4ka [110]

Answer:

:0

Explanation:

One second after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.Jun 17, 2017

6 0

3 years ago

Can u helmp me, i can give u a 25 points

tester [92]

They’re less reactive, they don’t react quickly with water or oxygen which they resist corrosion.

Answer : true

5 0

3 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$