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Effectus [21]
3 years ago
14

Insoluble substances can dissolve in all solvents. True or false

Chemistry
1 answer:
nadezda [96]3 years ago
7 0
I think it would be false
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How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?
larisa [96]

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>

<em />

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>

7 0
3 years ago
Read 2 more answers
Match: Substance ....... Definition Lewis acid A. Provides H+ in water Bronsted-Lowry base B. Provides OH - in water Arrhenius b
Igoryamba

Answer: F.    Electron pair acceptor

Explanation:

A Lewis acid can be properly defined as any substance such as  H+ (hygrogen ion) that can accept a pair of electron.

While a Lewis base is any substance such as (OH-) that can donate a pair of electron.

In the neutralization reaction between an acid ( H+ ) and a base (OH-). Hydrogen ion (H+ ) is the Lewis acid because it accepts an electron pair from (OH-).

Other examples of Lewis acid are;  Mg2+, K+

3 0
3 years ago
For the Earth's atmosphere, section number two in the pie graph BEST represents the percentage of A) argon B) oxygen C) nitrogen
larisa86 [58]

-<u><em>Oxygen</em></u>

According to Google these are the percentages of the <em>Earths Atmosphere</em>

<em>1</em> 78% - Nitrogen

<u>2</u> 21% - Oxygen

<em>3</em> 0.9% - Argon

<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.

3 0
3 years ago
Please help me with my regents practice :((
BlackZzzverrR [31]

Answer: 3 & 4

Explanation:

8 0
3 years ago
Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) =&gt; CH 3OH(l) Using the following information: C(graphite) +
Alika [10]

Answer:

\Delta H for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \Delta H_{1}= -393.5 kJ

2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l) \Delta H_{2}= (2\times -285.8)kJ

CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)  \Delta H_{3}= 726.4 kJ

---------------------------------------------------------------------------------------------------

C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ

4 0
3 years ago
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