Answer:
<h3>

</h3>
Explanation:
First balance the chemical equation:
⇄ 
two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container
Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp
![K_p=[CO_2]](https://tex.z-dn.net/?f=K_p%3D%5BCO_2%5D)
![[CO_2]=p](https://tex.z-dn.net/?f=%5BCO_2%5D%3Dp)



The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.
<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)
</span><span>(B)hydrochloric acid + sodium hydroxide
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)
</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)
<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)
AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.
Answer:
Explanation:
For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..
<span>n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9
First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4.
We know that 1dm3=1L, so H2SO4's molarity is
C=nV=18.0moles1.0L=18M
In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so
18.0moles1Lâ‹…98.0g1mole=1764g1L
Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution
98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolution→
masssolution=1764gâ‹…100.0g98g=1800g
Therefore, 1L of 98wt% H2SO4 solution will have a density of
Ď=mV=1800g1.0â‹…103mL=1.8gmL
H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be
cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m
Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that
100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4
100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O
So, H2SO4's mole fraction is
molefractionH2SO4=11+0.11=0.9</span>