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german
3 years ago
14

If r⃗ =bt2i^+ct3j^, where b and c are positive constants, when does the velocity vector make an angle of 45.0∘ with the x- and y

-axes?
Chemistry
1 answer:
Papessa [141]3 years ago
5 0
The displacement function is given by : r = bt2i + ct3j meters
The velocity function is the derivative of the displacement:
v = r' = 2bti + 3ct2j meter / second

Now, for <span> the velocity vector make an angle of 45.0∘ with the x- and y-axes, the i and j components have to be equal:
</span>2bt = 3ct2 (Now solve for t)
2bt = t (3ct)
either t = zero (rejected)
or t = 2b / 3c seconds (accepted)
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Can someone help balance this?
FromTheMoon [43]

Explanation:

2Fe+3Br---> 2FeBr3

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8 0
3 years ago
gThe mole fraction of potassium nitrate in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. What is the mol
xxMikexx [17]

Answer:

0.595 M

Explanation:

The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.

Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water

0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

0.0194x + 1.08 = x

x - 0.0194x = 1.08

0.9806x= 1.08

x= 1.08/0.9806

x= 1.1 moles of KNO3

Mole fraction of water= 55.6/1.1 + 55.6 = 0.981

If

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101

Molarity= 20.6/34.6

Molarity of KNO3= 0.595 M

8 0
3 years ago
For the chemical reaction CaI2+2AgNO3-&gt; 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i
Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

6 0
3 years ago
What is the mass of 89.6 L of SO2 gas at STP?
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