A bottle contains 158.1 grams of lead (II) sulfate (PbSO4), used for car batteries. How many moles does it contain?

The Al2O3 crystal structure (corundum) consists of an HCP arrangement of O2- ions; the Al3 ions occupy octahedral positions. Wha

ddd [48]

Answer:

2/3

Explanation:

Crystals structures can also be seen when two elements combines together and the perfect example is Al₂O₃ which is given in the question above. Just like it is given in the question above, the kind of arrangement in the crystal structure for Al₂O₃ is called HCP which stands for Hexagonally Closed Pack.

The aluminum ions which is in form of Al³⁺ occupies the two-third[2/3] positions while the position that the oxygen ion occupies is one[1].

8 0

2 years ago

PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!

Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)$

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

This is acid - base type reaction where

$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

5

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)$

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

7 0

2 years ago

Which of the following statements about minerals and ores is true?

Monica [59]

Answer:

please give a photo I can't now to help u

3 0

1 year ago

How does physical energy cause a rain puddle to disappear on a sunny day

goblinko [34]

The sun makes the moisture go into the air and evaporate

5 0

2 years ago

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to det

ivolga24 [154]

<u>Answer:</u> The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

<u>Explanation:</u>

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

6 0

2 years ago

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