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How many neutrons are in calcium nitride

xxTIMURxx [149]

Answer:

20

Explanation:

8 0

4 years ago

Electrons are to bonding as neutrons are to-

-BARSIC- [3]

I think it’s b sorry if it isn’t it’s been a while since I’ve done this but it should be positive charge

4 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?

Alja [10]

Answer:

$\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

$q= mc\Delta T$

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.

q= 47.1 J
m= 14.0 g
ΔT= 1.80 °C

Substitute these values into the formula.

$47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)$

Multiply the 2 numbers in parentheses on the right side of the equation.

$47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c$

$47.1 \ J = (25.2 \ g*\textdegree C) *c$

We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}$

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c$

$1.869047619 \ J/g *\textdegree C = c$

The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

$1.87 \ J/ g * \textdegree C =c$

The heat capacity of the liquid is approximately 1.87 J/g°C.

3 0

3 years ago

Which element in each of the following sets would you expect to have the lowest IE₃?

AysviL [449]

Answer:

(a) AL

(b) Sc

(c)Al

Explanation:

Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.

The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

3 0

3 years ago

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