4) the ice has to be heated up to above 0c first, as it is continuing to cool the water back toward 0c
5) 0.0338 kcal
(33.8/1000)
6) 1.2506 kcal
((33.8*37.0)/1000)
Answer:
The answer to the question is
The mass of the water in the container, assuming that all the heat lost by the copper is gained by the water is 116.23 grams
Explanation:
To solve this we list out the variables thus,
mass of copper = 110g
specific heat capacity of copper = 0.20 j/C g
Initial temperature of copper = 82.4 C
Final temperature of copper and water = 24.9 C
therefore
From the formula for sensible heat, ΔH = m×c×ΔT
we have, where
H = senced heat,
m = mass,
ΔT = temperature change
110 × 0.20 × (82.4 - 24.9) = mw × 4.186 × (24.9 - 22.3)
1265 J = mw ×10.8836 or mw = 116.23 g
Therefpre the mass of water is 116.23 g
Answer:
34.05dm^3 of nitrogen gas
Explanation:
First things first, we need to find the number of moles of Sodium azide. We can do that by using the formula m=nM, mass = no. moles x Molar Mass
Rearrange to solve for no. moles and substituting in the known values and we have:
n = m/M
no. moles = 130.0g / (2x(22.99+3x14.01))
no. moles = 130.0/130.0 (4.s.f.)
no. moles = 1
Now we can use the ratio given in the equation to find the number of moles of Nitrogen that will be made:
1 x 3/2 = 1.5 moles of Nitrogen
Now we use the constant that 1 mole of any gas will always have a volume of 22.7dm^3 at STP.
1.5 x 22.7 = 34.05dm^3 of nitrogen gas.
Hope this helped!
Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
<em />
Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
The molecular formula:
![N_x(O_2)_x](https://tex.z-dn.net/?f=N_x%28O_2%29_x)
![M_{N_x(O_2)_x}=x \times M_N + x \times 2 \times M_O \\ \\ m_N=14 \ \frac{g}{mol} \\ m_O=16 \ \frac{g}{mol} \\ M_{N_x(O_2)_x}=92 \ \frac{g}{mol} \\ \\ 92=x \times 14+x \times 2 \times 16 \\ 92=14x+32x \\ 92=46x \\ \frac{92}{46}=x \\ x=2 \\ \\ N_2(O_2)_2 \Rightarrow N_2O_4](https://tex.z-dn.net/?f=M_%7BN_x%28O_2%29_x%7D%3Dx%20%5Ctimes%20M_N%20%2B%20x%20%5Ctimes%202%20%5Ctimes%20M_O%20%5C%5C%20%5C%5C%0Am_N%3D14%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%0Am_O%3D16%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%20M_%7BN_x%28O_2%29_x%7D%3D92%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%20%5C%5C%0A92%3Dx%20%5Ctimes%2014%2Bx%20%5Ctimes%202%20%5Ctimes%2016%20%5C%5C%0A92%3D14x%2B32x%20%5C%5C%0A92%3D46x%20%5C%5C%0A%5Cfrac%7B92%7D%7B46%7D%3Dx%20%5C%5C%0Ax%3D2%20%5C%5C%20%5C%5C%0AN_2%28O_2%29_2%20%5CRightarrow%20N_2O_4)
The molecular formula is N₂O₄.