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Black_prince [1.1K]
3 years ago
15

Why are alkali metals more reactive than alkaline earth metals?

Chemistry
1 answer:
Neko [114]3 years ago
5 0
A: It takes more energy to remove two valence electrons from an atom than one valence electron. This makes alkaline Earth metals with their two valence electrons less reactive than alkali metals with their one valence electron.
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Determine the [OH−] , Ph, and POH of a solution with a [H+] of 0. 00017 m at 25 °C.
tino4ka555 [31]

Main Answer:

Given

pH = -log[H+]

= -log[0.00017]

= 3.769

We know that

pW = pH + pOH

and pW =14

pOH = 14-pH

=14-3.769

=10.231

According to the definition

pOH = -log[OH-]

10.231 = log[OH-]-1

[OH-]= 5.87 x 10-11

Explanation:

What is pH?

pH is defined as the concentration of H+ ion in the solution. If the pH value is less than 7, then the solution will be acidic. If the pH value is greater than 7, then the solution will be basic.

To know more about pH, please visit:

brainly.com/question/8758541

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4 0
2 years ago
And non-flammable gases<br> Noble gases are<br> that have low chemical
gtnhenbr [62]

Answer:

  • Noble gases are <u>odorless, colorless,</u> and nonflammable gases that have low chemical <u>reactivity</u><u>.</u>
  • The full <u>valence electron shells</u><u> </u>of these atoms make noble gases extremely <u>stable</u><u>.</u>
  • & they are <u>unlikely to form chemical bonds</u><u> </u>because they have little tendency to gain or lose

<u>electrons.</u>

5 0
3 years ago
in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,
insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, H_{supplied} = 3.34 × 10³ J

The amount thermal energy gained by the kettle, H_{absorbed}  = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}}  \times 100

Therefore, we get;

\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3}  \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
Por que esmas dificil erradicar la mineria ilegal q la mineria de grande empresas
kari74 [83]

Answer:

I Don't Understand

Explanation:

6 0
3 years ago
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