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labwork [276]
3 years ago
11

What practical use does adding salt to water have

Chemistry
1 answer:
igor_vitrenko [27]3 years ago
6 0
D. melting ice on roads and sidewalks, i think
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Which of the following elements is the smallest?
Oliga [24]

Answer:

Explanation:

The atomic radius of elements are used to estimate the sizes of elements. The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.

To solve this problem we will obtain the atomic radius values of the given elements from a standard atomic radius table;

          Si          111 pm

           P          98 pm

          Cl          79 pm

            S           87pm

pm = picometer

We see that chlorine has the least atomic radius

7 0
4 years ago
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Which of the following molecules has one lone pair of electrons?
BigorU [14]
You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.
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3 years ago
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What is the molar mass of BaBr2?<br><br>multiple choice on the picture
laiz [17]

B. 297.1 g/mol

Is your answer! ;)

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3 years ago
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What noble gas has the same electron configuration as the oxide ion?
Tems11 [23]
I love these type of questions :)

Final Answer: Neon
5 0
3 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
3 years ago
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