Answer: The momentum of the child and milk together is 58.125 kg.m/s
Explanation:
Momentum is defined as the product of object's mass and velocity.
Mathematically,

where, p = momentum
m = mass of the object
v = velocity of the object
In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:
....(1)
Given:

Putting values in equation 1, we get:

Hence, the momentum of the child and milk together is 58.125 kg.m/s
The solution is 45.7 % (NH₄)₂CO₃ by mass.
Mass of solution = 0.0332 kg + 0.0395 kg = 0.0727 kg
% (NH₄)₂CO₃ = Mass of (NH₄)₂CO₃/Total mass × 100 % = 0.0332 kg/0.0727 kg × 100 % = 45.7 %
Answer:125.84g
Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.
Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.
From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.
So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g
Hi
The volume is 58.1
The moler mass is 10.8
Molar mass details
10.81B (1*10.81)
Hope this helps
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.