Heat required = 173 kJ
<h3>Further explanation</h3>
Given
56.0 g of ice
Temperatur at 263 K(-10 C) to 400 K(127 C)
Required
Heat needed
Solution
1. raise the temperature(-10 C to 0 C)⇒c ice=2.09 J/g C

2. phase change (ice to water)⇒Heat of fusion water=334 J/g

3. raise the temperature(0 C to 100 C)⇒c water= 4.18 J/g C

4. phase change(water to vapor)⇒heat of vaporization water=2260 J/g

5. raise the temperature(100 C to 127 C)⇒c vapor=2.09 J/g C

Total heat :
1170.4+18704+23408+126560+3160.08=173,002.48 J=173 kJ
Answer:
K.E = 25 J
Explanation:
Given data:
Mass of ball = 0.5 g
Velocity of ball = 10 m/s
Kinetic energy = ?
Solution:
Formula:
K.E = 1/2 mv²
m = mass
v = velocity
Now we will put the values in formula.
K.E = 1/2 mv²
K.E = 1/2× 0.5kg × (10m/s)²
K.E = 1/2 ×0.5kg × 100m²/s²
K.E = 25 Kg.m²/s²
Kg.m²/s² = J
K.E = 25 J
Answer:
The salt that is produces is NaBr.
Explanation:
When an acid react with base it form the salt and water. The reaction is also called neutralization reaction because both neutralize each other.
In neutralization reaction equal amount of acid and base react to neutralize each other and equal amount of water and salt are formed. When pH does not reach to 7 its means there is less amount of one of reactant which is not fully neutralize.
Neutralization reactions are also used as first aid. For example when someone is dealing with HCl for cleaning purpose of toilet and get touched. It is advised to neutralize it with soap, milk or egg white.
Example:
Hydrobromic acid when react with the sodium hydroxide, a salt sodium bromide and water are formed.
Chemical equation:
HBr + NaOH → NaBr + H₂O
Properties of Sodium Bromide:
- It is present in the form of colorless crystals.
- It is also present in powder form.
- It is very toxic.
- It is soluble in water.
- It is bitter in taste and also odorless.
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
Answer:
Charles's law states that V1 / T1 = V2 / T2. However, you must make sure that T is in Kelvin, not C.
1.19E6 / (11 + 273) = V2 / (113 + 273)
V2 = 1.62E6
Explanation:
Gases expand when heated if the container they are in is flexible. When the gas is heated its molecules move faster and faster. The collisions of the gas particles with the flexible container wall expand.