<span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water.
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100.
One mole of Calcium Nitrate produces one mole of Calcium Carbonate.
i.e. 164 gms will produce 100gms of precipitate
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
The amount of heat energy added to silver to heat it from 25°C to 100°C is :
8737.5 J
<u>Given data: </u>
mass of silver ( m ) = 500 g
T1 = 25°C
T2 = 100°C
s ( specific heat of silver ) = 0.233 J/g.c
<h3 /><h3>Determine the amount of heat required </h3>
Applying the formula below for heat ( Q )
Q = ms * ΔT
= 500 * 0.233 * ( 100 - 25 )
= 8737.5 J
Hence we can conclude that the The amount of heat energy added to silver to heat it from 25°C to 100°C is : 8737.5 J.
Learn more about heat energy : brainly.com/question/13439286
Answer:
Ammonia and Hydrogen Chloride.
Explanation:
Assuming this is what you're referring to, Ammonia (NH3) and Hydrogen Chloride(NCI3). The concentrated ammonia (NH3) is placed on a pad in one end of a tube and hydrochloric acid (NCI3) on a pad at the other. Shortly after the gases will begin diffusing far enough to meet, a ring of solid ammonium chloride (NH4Cl) will be formed.
Hello!
To solve this problem we'll use the
Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=
4,19Now, we apply the equation as follows:
![pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BC_6H_5COONa%5D%7D%7B%5BC_6H_5COOH%5D%7D%20%29%3D4%2C19%2Blog%28%20%5Cfrac%7B0%2C15M%7D%7B0%2C25M%7D%20%29%3D3%2C97%20)
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is
3,97Have a nice day!
Answer:
0.583 kilojoules
Explanation:
The amount of heat required to pop a single kernel can be calculated using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of water (g)
c = specific heat capacity of water (4.184 J/g°C)
∆T = change in temperature
From the given information, m = 0.905 g, initial temperature (room temperature) = 21°C , final temperature = 175°C, Q = ?
Q = m × c × ∆T
Q = 0.905 × 4.184 × (175°C - 21°C)
Q = 3.786 × 154
Q = 583.044 Joules
In kilojoules i.e. we divide by 1000, the amount of heat is:
= 583.04/1000
= 0.583 kilojoules