Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. ... The equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of light.
B trench
Explanation:
The feature found at most convergent margins is a trench.
A trench is a large depression that typifies most convergent margins.
- Oceanic trenches are natural topographic depressions found on the sea floor.
- These depressions forms where two plate boundaries converges.
- The denser one slides beneath the less dense one into the asthenosphere below.
- This is called a subduction zone.
- The margin between the two plates that are depression is a trench.
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Answer:
2 C₄H₁₀(l) + 13 O₂(g) ⇄ 8 CO₂(g) + 10 H₂O(g)
Explanation:
When a substance burns we talk about a combustion reaction. When combustion is complete the products are carbon dioxide and water, like in this case. The equation is:
C₄H₁₀(l) + O₂(g) ⇄ CO₂(g) + H₂O(g)
First, we balance the element with the largest stoichiometric coefficient (C).
C₄H₁₀(l) + O₂(g) ⇄ 4 CO₂(g) + H₂O(g)
Then, we balance H because it is in just 1 compound on each side.
C₄H₁₀(l) + O₂(g) ⇄ 4 CO₂(g) + 5 H₂O(g)
Finally, we balance O.
C₄H₁₀(l) + 6.5 O₂(g) ⇄ 4 CO₂(g) + 5 H₂O(g)
Since we want the smallest whole numbers, we multiply all coefficients by 2.
2 C₄H₁₀(l) + 13 O₂(g) ⇄ 8 CO₂(g) + 10 H₂O(g)
