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Citrus2011 [14]

3 years ago

12

Atoms of the most reactive elements tend to have

2 answers:

kaheart [24]3 years ago

4 0

One or seven valance electons

Zolol [24]3 years ago

3 0

Answer choices:

A. one or seven valence electrons

B. eight valence electrons

C. four or five valence electrons

D. no valence electrons.

The answer is most likely A. one or seven valence electrons

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5. At 20°C, the water autoionization constant, Kw, is 6.8 ´ 10–15. What is the H3O+ concentration in neutral water at this tempe

rjkz [21]

Explanation:

Let us assume that the concentration of [ $OH^{-}$ and $H^{+}$ is equal to x. Then expression for $K_{w}$ for the given reaction is as follows.

$K_{w} = [OH^{-}][H^{+}]$

$K_{w} = x^{2}$

$6.8 \times 10^{-15} = x^{2}$

Now, we will take square root on both the sides as follows.

$\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}$

$[H^{+}] = 8.2 \times 10^{-8}$ M

Thus, we can conclude that the $H_{3}O^{+}$ concentration in neutral water at this temperature is $8.2 \times 10^{-8}$ M.

8 0

3 years ago

Read 2 more answers

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

anyanavicka [17]

Answer: The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

$(\alpha)=0.089$

percent ionisation = $0.089\times 100=8.9\%$

8 0

3 years ago

Which type of food is easily contaminated by bacteria?

Ivenika [448]

Answer:

frozen foods

Explanation:

explanation

7 0

3 years ago

Read 2 more answers

You are trying to determine the specific heat of a metal. You heat the 97 g piece of metal to 100 °C and place it in a calorimet

Brut [27]

Answer : The specific heat of the metal is, $0.658J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of metal = 97 g

$m_2$ = mass of water = 122 g

$T_f$ = final temperature of mixture = $28.9^oC$

$T_1$ = initial temperature of metal = $100^oC$

$T_2$ = initial temperature of water = $20.0^oC$

Now put all the given values in the above formula, we get

$97g\times c_1\times (28.9-100)^oC=-122g\times 4.18J/g^oC\times (28.9-20.0)^oC$

$c_1=0.658J/g^oC$

Therefore, the specific heat of the metal is, $0.658J/g^oC$

8 0

4 years ago

Given a certain quantity of reactant, you calculate that a particular

zlopas [31]

Answer:

Percent yield = 114.5%

Explanation:

Given data:

Calculated amount of product/ theoretical yield = 55 g

Actual amount of product after performing reaction = 63 g

Percent yield = ?

Solution:

Formula:

Percent yield = ( actual yield / theoretical yield )× 100

by putting values,

Percent yield = (63 g/ 55 g)× 100

Percent yield = 114.5%

4 0

3 years ago