Answer : The metal used was iron (the specific heat capacity is
).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of unknown metal = ?
= specific heat of water = 
= mass of unknown metal = 150 g
= mass of water = 200 g
= final temperature of water = 
= initial temperature of unknown metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get


Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).
Therefore, the metal used was iron (the specific heat capacity is
).
44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3
46. H Br
δ+ δ−
48. The metallic potassium atoms lose one electron and form +1 cations,
and the nonmetallic fluorine atoms gain one electron and form –1 anions.
K → K+
+ e–
19p/19e–
19p/18e–
F + e–
→ F–
9p/9e–
9p/10e–
The ionic bonds are the attractions between K+
cations and F–
anions.
50. See Figure 3.6.
52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal
54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic
56. (a) 7 (b) 4
58. Each of the following answers is based on the assumption that nonmetallic
atoms tend to form covalent bonds in order to get an octet (8) of
electrons around each atom, like the very stable noble gases (other than
helium). Covalent bonds (represented by lines in Lewis structures) and lone
pairs each contribute two electrons to the octet.
(a) oxygen, O
If oxygen atoms form two covalent bonds, they will have an octet of electrons
around them. Water is an example:
H O H
(b) fluorine, F
If fluorine atoms form one covalent bond, they will have an octet of electrons
around them. Hydrogen fluoride, HF, is an example:
H F
(c) carbon, C
If carbon atoms form four covalent bonds, they will have an octet of electrons
around them. Methane, CH4, is an example:
H H
H
H
C
(d) phosphorus, P
If phosphorus atoms form three covalent bonds, they will have an octet
Answer:
longitudinal wave
Explanation:
hope it helps you and Mark me down as brainlist
Answer:
The value is 
Explanation:
From the question we are told that

The initial volume of the fluorocarbon gas is 
The final volume of the fluorocarbon gas is
The initial temperature of the fluorocarbon gas is 
The final temperature of the fluorocarbon gas is 
The initial pressure is 
The final pressure is 
Generally the equation for adiabatically reversible expansion is mathematically represented as
![T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }](https://tex.z-dn.net/?f=T_2%20%3D%20%20T_1%20%20%2A%20%5B%20%5Cfrac%7BV_1%7D%7BV_2%7D%20%5D%5E%7B%5Cfrac%7BR%7D%7BC_v%7D%20%7D)
Here R is the ideal gas constant with the value

So
=> 
Generally adiabatic reversible expansion can also be mathematically expressed as

=>
=> 
=>
So

=> 
Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"