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kap26 [50]
2 years ago
8

How much momentum does a 200 kg rhino have that is running at 35 m/s E

Chemistry
2 answers:
nirvana33 [79]2 years ago
5 0

Answer:

7000 kg*m/s E

Explanation:

Momentum formula: p=mv

m=200kg

v=35 m/s East

p=(200kg)(35m/s E)

m=7000 kg*m/s E

If you want to simplify it further, m=7*10^3 kg*m/s E

AlexFokin [52]2 years ago
4 0

Answer:

\boxed {\boxed {\sf 7000 \ kg*m/s \ E}}

Explanation:

Momentum is the product of mass and velocity.

p=m*v

The mass is 200 kilograms and the velocity is 35 meters per second east.

m= 200 \ kg \\v= 35 \ m/s \ E

Substitute the values into the formula.

p= 200 \ kg * 35 \ m/s \ E

Multiply.

p= 7000 \ kg * m/s \ E

The rhino has 7000 kg*m/s E of momentum.

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Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.
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The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

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Relative atomic mass data from a modern periodic table:

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\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}],

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of \rm H_2 that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be \rm MgCl_2.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq).

Balance the equation. \rm MgCl_2 contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq).

The number of \rm Mg and \rm Cl atoms shall be the same on both sides. Therefore

\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq).

The number of \rm H atoms shall also conserve. Hence the equation:

\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq).

How many moles of HCl are available?

M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}.

\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol.

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of \rm HCl is 2 while the coefficient in front of \rm H_2 is 1. In other words, it will take two moles of \rm HCl to produce one mole of \rm H_2. \rm 4.52576\;mol of \rm HCl will produce only one half as much \rm H_2.

Alternatively, consider the ratio between the coefficient in front of \rm H_2 and \rm HCl is:

\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}.

\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol.

What will be the volume of that many hydrogen gas?

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V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L.

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