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3 years ago

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Which elements are included in group 5a

1 answer:

RUDIKE [14]3 years ago

7 0

Answer:

Group 5A (or VA) of the periodic table are the pnictogens:

the nonmetals nitrogen (N),

and phosphorus (P),

the metalloids arsenic (As)

antimony (Sb),

the metal bismuth (Bi).

and moscovium (Mc).

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3 years ago

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Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a uni

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Answer : The mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Explanation : Given,

Number of atoms of $V_2O_5$ = $1.0\times 10^{6}$

Molar mass of $V_2O_5$ = 181.88 g/mole

In $V_2O_5$ , there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of $V_2O_5$ .

As, $2\times 6.022\times 10^{23}$ number of vanadium atom present in 1 moles of $V_2O_5$

So, $1.0\times 10^{6}$ number of vanadium atom present in $\frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19}$ moles of $V_2O_5$

Now we have to determine the mass of $V_2O_5$ .

$\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5$

$\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g$

Therefore, the mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

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4 years ago

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Determine the partial negative charge on the bromine atom in a c−br bond. the bond length is 1.93 å and the bond dipole moment i

vaieri [72.5K]

Answer:

The value is $x = 0.151 \ e$

Explanation:

From the question we are told that

The bond length is $l = 1.93\ \r a = 1.93 *1 *10^{-10} =1.93 *10^{-10}\ m$

The bond dipole moment is $\mu = 1.40 d = 1.40 * 3.33564 *10^{-30} = 4.6699 *10^{-30} \ C \cdot m$

Generally the dipole moment is mathematically represented as

$\mu = Q * l$

Here Q is the partial negative charge on the bromine atom

So

$Q = \frac{\mu}{ l}$

=> $Q = \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }$

=> $Q = 2.42 *10^{-20} C$

Generally

1 electronic charge(e) is equivalent to $1.60*10^{-19} C$

So x electronic charge(e) is equivalent to $Q = 2.42 *10^{-20} C$

=> $x = \frac{2.42 *10^{-20}}{1.60*10^{-19} }$

=> $x = 0.151 \ e$

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3 years ago