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Alborosie
3 years ago
7

An unknown element, x, reacts with rubidium to form the compound rb2x. in other compounds this element also can accommodate up t

o 12 electrons rather than the usual octet. what element could x be?
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

There are few elements that can form 12 electron rather than the usual 8. Therefore these are the possible answers:

Sulfur = 2Na + S --> Na2S <span>
Selenium = 2Na + Se --> Na2Se 
<span>Tellurium = 2Na + Te --> Na2Te</span></span>

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A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c
oksano4ka [1.4K]

Answer: 0.17M

Explanation:

The equation for the reaction is :

2HNO3 + Ba(OH)2 —> Ba(NO3)2 + 2H2O

From the balanced equation, we obtain :

nA = mole of acid = 2

nB = mole of the base = 1

From the question, we obtain:

Va = Vol. Of acid = 0.105L

Ma = conc. Of acid =?

Vb = Vol of base = 35.7 mL = 0.0357L

Mb = conc. of base = 0.25M.

We solve for the conc. of the acid using:

MaVa / Mb Vb = nA / nB

(Ma x 0.105) / (0.25x0.0357) = 2

Cross multiply to express in linear form. We have:

Ma x 0.105 = 0.25 x 0.0357 x 2

Divide both side by 0.105. We have

Ma = (0.25 x 0.0357 x 2) / 0.105

Ma = 0.17M

Therefore the concentration of the acid(HNO3) is 0.17M

3 0
3 years ago
PLEASE HELP!!! ON A TIMER, WILL MARK BRAINLIEST
Nadusha1986 [10]

Answer:

<em>Well, Your best answer will be is  2H+ + 2OH-  ->  2H2O but you have to reduce it to H+ + OH-  ->  H2O. </em><em>Good Luck!</em>

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5 0
3 years ago
A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so
uranmaximum [27]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>

Explanation:

[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>

Ionic equation:

{ \bf{2OH { }^{ - }  _{(aq)} + 2H { }^{ + } _{(aq)}→  2H _{2} O _{(l)} }}

first, we get moles of potassium hydroxide in 28.94 ml :

{ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

{ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\  = { \underline{423.95 \: g \approx424 \: grams}}

for the molar mass:

8 0
3 years ago
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