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3 years ago

What is the stable configuration of the nitride ion (N^-3)

1 answer:

3 0

1s^2 ^2s^2 2p^6

Explanation:
Nitride is just an N3− ion. We know this because the -ide- suffex indicates a mono atomic anion, and we assume that the anion will have enough additional electrons to achieve a noble gas configuration (nitrogen is 3 electrons short of a full octet).

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2.0 L of a gas is at STP. What is the new pressure in kPa if the volume changes to 10.5 L and the temperature and # of molecules

Alexandra [31]

Answer:

What is the new volume if the temperature is constant? V=2.50L. P = lookPa. P2=40k Pa. V2 = x. PV = P2 ... If a sample of gas occupies 6.8 L at 327°C

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3 years ago

Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O

Bess [88]

40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O

5 0

3 years ago

What is the genetic makeup of both parents ?

Effectus [21]

Answer:

A, T, C y G, son las "letras" del código del ADN; representan los compuestos químicos adenina (A), timina (T), citosina (C) y guanina (G), respectivamente, que constituyen las bases de nucleótidos del ADN. ...El código genético es el conjunto de reglas que define cómo se traduce una secuencia de nucleótidos en el ARNm a una secuencia de aminoácidos en una proteína

7 0

3 years ago

Hydrogen is the only atom that does not have ____ electrons in the 1st energy level/shell.

dalvyx [7]

Hydrogen is the only atom that does not have neutron electrons in the first energy level/shell.

4 0

3 years ago

We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a

sukhopar [10]

Answer:

See Explanation

Explanation:

$\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n$

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

= 236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) $\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy$

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0

3 years ago