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aleksandrvk [35]
3 years ago
12

Nitrogen can ionically bond with an unknown element X from group 2. How many ions of element X are required to create a stable i

onic compound with nitrogen?
Question 15 options:


2


03


4


There is not enough information
Chemistry
2 answers:
Nataliya [291]3 years ago
8 0
I can tell you there certainly is enough information, so we can eliminate the fourth option right away.
Group two elements tend to form 2+ cations, and nitrogen forms 3- anions. To make a stable substance, we need those to cancel out. Calling the group 2 element X:
N2X3 would cause the charge to be 2(-3) + 3(2) = 0
This would mean 3 ions of X and 2 anions of N would be a stable ionic compound.
torisob [31]3 years ago
3 0

Answer:

B.) 3

Explanation:

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gamma

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A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem
Leto [7]

Answer : The final temperature of the copper is, 87.29^oC

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat gained  = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = 0.0923cal/g^oC      

\Delta T=\text{Change in temperature} 

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25^oC

Now put all the given values in the above formula, we get the final temperature of copper.

299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)

T_{final}=87.29^oC

Therefore, the final temperature of the copper is, 87.29^oC

4 0
3 years ago
6) A volume of 473 mL of oxygen was collected at 27°C. What volume would the oxygen occupy at
sertanlavr [38]

Answer : The volume of oxygen occupy at 173° would be, 703.2 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=473mL\\T_1=27^oC=(27+273)K=300K\\V_2=?\\T_2=173^oC=(173+273)K=446K

Now put all the given values in above equation, we get:

\frac{473mL}{300K}=\frac{V_2}{446K}\\\\V_2=703.2mL

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL

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