Answer:
The answer will be 2.98K
Explanation:
Using the formula:
Q = mc∆T
Q= 5,800 (heat in joules)
m= convert 15.2kg to g which is 15200g (mass in grams)
c= 0.128 J/g °c (Specific heat capacity)
∆T= what we need to find (temperature change)
5800J = 15200g x 0.128 x ∆T
= 2.98K
C=0.10 mol/l
pH=-lg[H⁺]
HCl = H⁺ + Cl⁻
pH=-lgc
pH=-lg0.10=1.0
pH=1.0
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
NaBrO3 is the chemical formula for Sodium Bromate.
