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natita [175]
3 years ago
13

ΚΕ: m.22 2 What is the kinetic energy of 10kg dog running with a velocity of 12m/s?

Chemistry
1 answer:
BigorU [14]3 years ago
3 0

Answer:

The answer is

<h2>720 Joules</h2>

Explanation:

The kinetic energy of a body can be found by using the formula

<h3>K.E =  \frac{1}{2}  {mv}^{2}</h3>

where

m is the mass

v is the velocity / speed

From the question

mass = 10 kg

velocity = 12 m/s

Substitute the values into the above formula and solve

That's

<h3>KE =  \frac{1}{2}  \times 10 \times  {12}^{2}  \\   = 5 \times 144  \:  \:</h3>

We have the final answer as

<h3>720 Joules</h3>

Hope this helps you

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of hea
SCORPION-xisa [38]
Heat energy can be calculated by using the specific heat of a substance multiplying it to the mass of the sample and the change in temperature. It is expressed as: 
Energy = mCΔT2520= 10.0(C) (70.0 - 10.0)C = 4.2 J/ kg K
7 0
3 years ago
Read 2 more answers
13. How many moles of zinc are in 25.00 g Zn?
topjm [15]
0.3824 moles of Zinc
4 0
3 years ago
Please help it’s already overdue
Gnesinka [82]

Answer:

no se amigo esquema no entiendo ingles

7 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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