Amount of SeO2-3 is -

The correct option is C 97 mgMilliequivalents of SeO^2 _3=55/36 Weight of SeO^2 _3= mEq. × 10^ 3× Equivalent weight (Ew of SeO ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Arrhenius Acid

Amount of S...

Question

Amount of $S e O_{3}^{2 -}$ is :

19.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

194

No worries! We‘ve got your back. Try BYJU‘S free classes today!

970

No worries! We‘ve got your back. Try BYJU‘S free classes today!

97

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $97$ mg
Milliequivalents of $S e O_{3}^{2 -} = \frac{55}{36}$

Weight of $S e O_{3}^{2 -} =$ mEq. $\times 10^{- 3} \times$ Equivalent weight $(E w$ of $S e O_{3}^{2 -} = \frac{127}{2} (n = 2))$
$= \frac{55}{36} \times 10^{- 3} \times \frac{127}{2}$ g $= 0.097$ g $= 97$ mg

Suggest Corrections

Similar questions

A water contains the following dissolved ions.
$[N a^{+}] = 56 m g / L; [C a^{2 +}] = 40 m g / L;$
$[M g^{2 +}] = 30 m g / L; [A l^{3 +}] = 3 m g / L;$
$[H C O_{3}^{-}] = 190 m g / L; [C l^{-}] = 165 m g / L;$
Water $p H$ is $7$ .
Atomic weights:
$C a = 40, M g = 24, A l = 27, H = 1, C = 12, O = 16, N a = 23, C l = 35.5$

The non-carbonate hardness of the sample in $m g / L$ as $C a C O_{3}$ is

M g + (N O_{3})^{-} \to

M g + (S O_{4})^{2 -} \to

M g + (P O_{4})^{3 -} \to

Q. Electrode potential for Mg electrode varies according to the equation

E_{M} g^{2} + | M g

E_{M} g^{2} + | M g^{Θ}

\frac{0.059}{2} l o g \frac{1}{[M g^{2 +}]}

. The graph of

E_{M g^{2 +} | M g}

l o g [M g^{2 +}]

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.3, approximately 108) the amount of silver deposited is:

1g + 300 mg = ___.

Join BYJU'S Learning Program

Amount of SeO2−3 is :

The correct option is C 97 mgMilliequivalents of SeO2−3=5536Weight of SeO2−3= mEq. × 10−3× Equivalent weight(Ew of SeO2−3=1272(n=2))=5536×10−3×1272 g=0.097 g =97 mg

Amount of $S e O_{3}^{2 -}$ is :

The correct option is C $97$ mg
Milliequivalents of $S e O_{3}^{2 -} = \frac{55}{36}$

Weight of $S e O_{3}^{2 -} =$ mEq. $\times 10^{- 3} \times$ Equivalent weight $(E w$ of $S e O_{3}^{2 -} = \frac{127}{2} (n = 2))$
$= \frac{55}{36} \times 10^{- 3} \times \frac{127}{2}$ g $= 0.097$ g $= 97$ mg