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2 years ago

Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.

Chemistry

1 answer:

Fittoniya [83]2 years ago

4 0

Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.

Group of answer choices

hydrochloric acid

sodium bicarbonate

calcium carbonate

sodium chloride

sodium hydroxide

Answer:

calcium carbonate

Explanation:

A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.

A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.

Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.

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Acids and bases can neutralize each other in double displacement reactions. For example, if hydrogen chloride (a strong acid, HC

scoray [572]

Answer:

The answer to your question is HCl + NaOH ⇒ NaCl + H₂O

Explanation:

Data

Double displacement reaction

Balanced chemical reaction

HCl + NaOH ⇒ NaCl + H₂O

Reactants Elements Products

1 Chlorine (Cl) 1

1 Sodium (Na) 1

2 Hydrogen (H) 2

1 Oxygen (0) 1

As we can see, the reaction is balanced and the coefficients of all reactants and products are 1, but the number is not written in a balanced reaction.

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3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

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3 years ago

A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a

lorasvet [3.4K]

For equal moles of gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K

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3 years ago

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3 years ago