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Yuri [45]
3 years ago
14

Which of the following statements concerning the density of a gas is true?

Chemistry
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

can u give us the options

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In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
The molecular weight of Calcium Carbonate, CaCO3, is 100.09g/mol
laila [671]

Answer:

True

Explanation:

Molecular mass of CaCo_3:

Molecular mass of Ca + molevular mass of C + molecular mass of O × 3

40g + 12g + 16 × 3g

= 40g + 60 = 100g

Thus its true

3 0
3 years ago
Which element has the greatest density at stp? 1. carbon 2. copper 3. chlorine 4. calcium
Vanyuwa [196]
Copper will have the greatest density, followed by calcium, another metal. Carbon will be the third most dense followed by chlorine, which, as a gas, will be MUCH less dense than the other 3
8 0
3 years ago
It has been an expense for the Australian government (taxpayers) to try to control the rabbit population
Eddi Din [679]
I believe it’s true hope this helps
7 0
3 years ago
Explain how to choose a proper indicator in an acid-base titration such as adding a strong acid (HCI) to a strong base (NaOH) ,
Jet001 [13]
The pH of the solution at equivalence point is dependent on the strength of the acid and strength of the base used in the titration.

For strong acid-strong base titration, pH = 7 at equivalence point
For weak acid-strong base titration, pH > 7 at equivalence point
For strong acid-weak base titration, pH < 7 at equivalence point

https://www.khanacademy.org/test-prep/mcat/chemical-processes/titrations-and-solubility-equilibria/a...  

The indicator changes color when the pH changes at the endpoint of the titration. So, you need to determine what is present at the point and find the pH at the point. Then, you can reference a table of indicators to choose one whose color will change over the pH that includes your equivalency point.
5 0
3 years ago
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