Answer:
12.33 cal/sec
Explanation:
As we know,
1 Kcal = 1000 cal
So,
0.74 Kcal = X cal
Solving for X,
X = (0.74 Kcal × 1000 cal) ÷ 1 Kcal
X = 740 cal
Also we know that,
1 Minute = 60 Seconds
Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,
= 740 cal / 60 sec
= 12.33 cal/sec
<u>Answer:</u> The correct answer is Option 1.
<u>Explanation:</u>
Neutralization reaction is defined as the reaction in which acid reacts with a base to produce a salt along with water.
Here, HCl is an acid and 
is a base. When these two compounds react, the salt obtained is calcium chloride.
The equation for the above reaction is given by:
Hence, the correct answer is Option 1.
Answer:
A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium. A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.
Answer:- There are 
moles.
Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is 
and it's molar mass is 294.31 grams per mole.
mg are converted to grams and then the grams are converted to moles as:
= moles of aspartame
So, there would be moles of aspartame in 1.00 mg of it.
Answer:
2= its color
Explanation:
Transition elements are present in the middle of periodic table. These are d-block elements.
These are 38 elements.
All transition elements have partially filled d orbitals.
They showed color in compound because of d-d transition.
During the d-d transition electron absorbed the energy and emit the reminder energy. The emission is usually in the form of color light.
The color of ion is complementary to the absorbed color.
The transition elements are used as a catalyst in industries such as polymer, petroleum industries.
They are ductile, conduct heat and electricity.
