Based on the data given, which factor had the greatest effect on the behavior of the gas

Read that From picture and tell me the answer please

Dafna11 [192]

Answer:

1g Hydrogen

Explanation:

<h3>Getting to the equation:</h3>

Calcium in water reacts vigorously to give a cloudy white Precipitate (compound) called Calcium hydroxide alongwith the evolution of Hydrogen gas.

$\boxed{ \mathsf{Ca + H_2O \rightarrow Ca(OH)_2 + H_2}}$

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<h3>Balancing the equation: </h3>

This reaction is not in it's balanced form! The number of atoms of Hydrogen on the left is 2 while that on the right is 4,I.e.,they're not equal.

Adding a 2 in front of H2O solves the problem by making the number of atoms of each element on both the sides equal.

$\mathsf{Ca +2 H_2O \rightarrow Ca(OH)_2 + H_2}$

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<h3>Observations:</h3>

Looking into the equation more carefully, we see:

1 atom of Calcium reacts with 2 molecules of water to give 1 molecule of Calcium Hydroxide alongwith 1 molecule of Hydrogen gas.

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<h3>Gram atomic and molecular masses </h3>

Mass of one atom of Calcium = it's gram atomic mass

= 40 g

Mass of one "molecule" of Hydrogen

= it's Gram molecular mass

= gram mass of one atom × number of atoms in one molecule

= 1 × 2

= 2 g

So,

according to our observation:

One atoms of Calcium gives one molecule of Hydrogen (during the particular reaction)

=> 40g of Calcium gives = 2g of Hydrogen

•°• 1 g of Calcium gives = $\frac{2}{40}$

= $\frac{1}{20}$ g Hydrogen

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<h3>Answer:</h3>

We're provided with 20g of Calcium,

=> 20g of Calcium gives = 20 × $\frac{1}{20}$ g H2

= 1 g H2

_______________

Hope this helps!

6 0

3 years ago

The volume of a fixed amount of gas is doubled, and the absolute temperature is doubled. According to the ideal gas law, how has

neonofarm [45]

Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

3 0

3 years ago

Sodium has______ Valence electrons and bismuth has ________ electrons

Triss [41]

Answer:

Sodium has 1 valence electron and bismuth has 83 electrons (5 valence electrons)

Explanation:

5 0

3 years ago

Cryogenics has the potential to be useful in a variety of fields, including medicine. Suppose you have engineered a method to su

tatiyna

Explanation:

It is known that the specific heat capacity of Liver $(C_{p})$ is 3.59 kJ $kg^{-1}.K^{-1}$

It is given that :

Initial temperature of Liver = Body temperature = $37^{o}C$ = 310 K

Final temperature of Liver = 180 K

Relation between heat energy, mass, and change in temperature is as follows.

Q = $m \times C_{p} \times \Delta T$

Now, putting the given values into the above formula as follows.

Q = $m \times C_{p} \times \Delta T$

Q = $1.5 kg \times 3.59 kJ/kg.K \times (310 - 180) K$

= 700.05 kJ

Therefore, we can conclude that amount of heat which must be removed from the liver is 700.05 kJ.

7 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

4 0

4 years ago