Calculating speed or atoms and molecules or planets!
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The answer to your question is A.
Answer:
1.5x10²² particulates
Explanation:
Assuming ideal behaviour, we can solve this problem by using the <em>PV=nRT </em>formula, where:
- V = 250 mL ⇒ 250 / 1000 = 0.250 L
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 15 °C ⇒ 15 + 273 = 288 K
We <u>input the given data</u>:
- 2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 K
And <u>solve for n</u>:
Finally we <u>calculate how many particulates are there in 0.025 moles</u>, using <em>Avogadro's number</em>:
- 0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulates
Answer:
Option C. Na⁺(aq) and Cl¯(aq)
Explanation:
From the question given above, we obtained the following:
Na₂SO₄(aq) + CaCl₂(aq) —> CaSO₄(s) + 2NaCl(aq)
Ionic Equation:
2Na⁺(aq) + SO₄²¯(aq) + Ca²⁺(aq) + 2Cl¯(aq) —> CaSO₄(s) + 2Na⁺(aq) + 2Cl¯(aq)
From the ionic equation above, we can see that Na⁺(aq) and Cl¯(aq) are present on both side of the equation.
Therefore, Na⁺(aq) and Cl¯(aq) are the spectator ions because they did not participate directly in the reaction.
Answer:
d
. The volume of the solvent used was less than 5 liters.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the volume of the stock (initial) solution by using the following equation:
Thus, we solve for, V1, which stands for the aforementioned volume of stock solution:
Then, we plug in to obtain:
Now, since the final volume was 5 L, we can infer that the volume of solvent is 4.5 L and that of the stock solution 0.5 L for a total of 5 L of diluted solution; therefore, the correct reasoning is d
. The volume of the solvent used was less than 5 liters.
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