Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose = 
x 55
= 
x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g
Answer:
2 half-lives=0.8
6 half-lives= 0.05
Explanation
Half-lives are constant and always decrease by half, implying that the concentration decreases by half at a consistent rate.
3.2/2= 1.6/2= 0.8 is two half-lives
3.2/2= 1.6/2= 0.8/2= 0.4/2= 0.2/2= 0.1/2=0.05 is six half-lives
(a) We know that work is the product of Force and Distance so: (in this
case Distance is negative since going down so –d)
work = force * distance
work = M * (g - g/4) * -d
work = -3Mgd/4 <span>
(b) The work by the weight of the block is simply:</span>
work = Mgd <span>
(c) The kinetic energy is simply equivalent to the
net work, therefore:</span>
KE = net work
KE = Mgd/4 <span>
(d) The velocity is:</span>
v = √(2*KE/M)
Plugging in the value of KE from c:
v = √(2*Mgd / 4M)
<span>v = √(gd / 2) </span>
I believe <span>erosion is what you are looking for..</span>
Equation is as follow,
<span> 2 AgNO</span>₃<span> + MgBr</span>₂<span> </span>→ <span>2 AgBr + Mg(NO</span>₃<span>)</span>₂
According to eq.
339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr
Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g
X = 24.87 g of AgBr
