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3 years ago

What is the momentum of a bird with a mass of 2 kg flying at 9 m/s? *

Chemistry

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 2 × 9

We have the final answer as

Hope this helps you

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Sally has touched a hot pan on the stove and burnt her hand. This example shows what type of heat transfer?

Minchanka [31]

Answer:

Transfer of Heat

You might have observed that a frying pan becomes hot when kept on a flame. It is because the heat passes from the flame to the utensil. When the pan is removed from the fire, it slowly cools down. Why does it cool down? The heat is transferred from the pan to the surroundings. So you can understand that in both cases, the heat flows from a hotter object to a colder object. In fact, in all cases heat flows from a hotter object to a colder object.

Explanation:

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6 0

3 years ago

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 numerical value of the equilibrium constant Kc?

Nezavi [6.7K]

ANSWER

EXPLANATION

Given that;

The number of moles of NH3 is 3 moles

The number of moles of H2 is 1 mole

The number of moles of N2 is 2 moles

At equilibrium, the concentration of ammonia is 1.4 moles/L

To find the value of Kc, follow the steps below

Step 1: Write the balanced equation of the reaction

$\text{ N}_{2(g)}\text{ + 3H}_{2(g)}\text{ }\rightleftarrows\text{ 2NH}_{3(g)}$

Step 2: Write the equation of the reaction in terms of Kc

$\text{ K}_C\text{ = }\frac{[\text{ NH}_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$

Step 3: Find the concentration of each reactant at equilibrium using a stoichiometry ratio

From the reaction above, you will see that 1 mole of Nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.

Let x represents the concentration of nitrogen at equilibrium

Recall, that the concentration of ammonia at equilibrium is 1.4 moles/L

$\begin{gathered} \text{ 2 mole NH}_3\text{ }\rightarrow\text{ 1 mole N}_2 \\ \text{ 1.04 mole/L NH}_3\text{ }\rightarrow\text{ x moles/L N}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ x moles/L N}_2\text{ = 1 mole N}_2\times1.4\text{ mol/L} \\ \text{ Isolate x} \\ \text{ x mol/L N}_2\text{ = }\frac{1\text{ moles N}_2\times1.41\cancel{\frac{mol}{L}}}{2\cancel{moles}} \\ \text{ x mol/L = }\frac{1.04}{2} \\ \text{ x= 0.52 mole/L} \end{gathered}$

Since the initial number of moles of nitrogen is 1 mole, hence, the concentration of nitrogen at equilibrium is calculated below as

Concentration at equilibrium = 1 -0.52

Concentration of nitrogen at equilibrium = 0.48 mole/L

The next step is to find the concentration of hydrogen at equilibrium

Let y represent the mole of hydrogen at equilibrium

$\begin{gathered} \text{ 2 moles NH}_3\rightarrow\text{ 3 moles H}_2 \\ \text{ 1.04 moles/L NH}_3\text{ }\rightarrow\text{ y moles/L H}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ y moles/L H}_2\text{ = 1.04 moles/L NH}_3\times3\text{ moles H}_2 \\ \text{ Isolate y} \\ \text{ y moles/L H}_2\text{ = }\frac{1.04\cancel{\frac{moles}{L}}NH_3\times3mole\text{ H}_2}{2\cancel{molsNH_3}} \\ y\text{ = }\frac{1.40\times3}{2} \\ \text{ y = }\frac{3.12}{2} \\ \text{ y = 1.56 moles} \end{gathered}$

Since the initial concentration of hydrogen is 2 moles, hence the concentration of hydrogen at equilibrium can be calculated below as

Concentration at equilibrium = 2 - 1.56

Concentration at equilibrium = 0.44 mole/L

Step 4: Find the value of Kc using the equation in step 2

$\text{ Kc = }\frac{[NH_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$ $\begin{gathered} \text{ Kc = }\frac{1.04^2}{0.48\times0.44^3} \\ \\ \text{ K}_c=\text{ }\frac{1.0816}{0.48\times\text{ 0.085184}} \\ \\ \text{ kc = }\frac{1.0816}{0.0408832} \\ \text{ kc = 26.4558547276} \end{gathered}$

Hence, the value of Kc is 26

6 0

1 year ago

I need help with my assignment asap

NeX [460]

1) taking drugs to prevent diseases

2) liver cirrhosis

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3 years ago

The rate constant for a particular zero-order reaction is 0.075 M s-1. If the initial concentration of reactant is 0.537 M it ta

Phoenix [80]

Answer : It takes time for the concentration to decrease to 0.100 M is, 22.4 s

Explanation :