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astra-53 [7]
3 years ago
11

Calcium is added to water,the liberated gas is​

Chemistry
2 answers:
vlabodo [156]3 years ago
4 0

Answer:

In both cases, the gas evolved is H2. When calcium reacts with water the heat evolved is not suffcient for hydrogen to catch fire. On the other hand, sodium metal reacts with water violently and in this case a lot of heat is evolved which is sufficient for hydrogen to catch fire.

Explanation:

may this answer is helpful for you

olchik [2.2K]3 years ago
3 0

Hydrogen

The reaction is given by

\boxed{\sf {Ca\atop Calcium}+{H_2O\atop Water}\longrightarrow {Ca(OH)_2\atop Calcium\:Hydroxide}+{H_2\atop Hydrogen}}

Balanced equation:-

\boxed{\sf {Ca\atop Calcium}+{2H_2O\atop Water}\longrightarrow {Ca(OH)_2\atop Calcium\:Hydroxide}+{H_2\atop Hydrogen}}

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How many moles of oxygen are required to react completely with 5 mol C8H18?
Nata [24]

Answer:

62.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.

Thus, 62.5 moles of O₂ is needed for the reaction.

6 0
3 years ago
You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f
dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

(1 atm = 760 mmHg)

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65  K

Putting values in above equation, we get:

(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole

Percentage recovery of carbon dioxide gas =  49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

\frac{49.4}{100}\times  0.0438 mol=0.02164 mol

2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

\frac{2}{1}\times 0.02164 mol=0.04328 mol

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0
3 years ago
What was Ernest Rutherford's experiment?
Dafna1 [17]

Answer:

C. He shot tiny alpha particles through a piece of gold foil.

Explanation:

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