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2 years ago

A group of students are measuring how the acceleration of a mass will change when it is acted

Chemistry

1 answer:

Virty [35]2 years ago

5 0

The scatter plot that shows a positive correlation between force and acceleration.

<h3>What is the relationship between acceleration of an object and the applied force?</h3>

The acceleration of an object is the change in velocity with time.

Acceleration = change in velocity/time

Force is defined as an agent which causes a change in the motion or state of rest of a body.

According to Newton's law of motion, the rate of change of velocity of an object is directly proportional to the applied force and takes place in the direction of the applied force.

Force and acceleration have a positive correlation.

Therefore, the scatter plot which will most closely match the measurements that the students will obtain is that which shows a positive correlation between force and acceleration.

In conclusion, acceleration of a object is proportional to the applied force.

Learn more about force and acceleration at: brainly.com/question/14343220

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The type of igneous rock which has a large number of cavities

lisabon 2012 [21]

The answer would be scoria I believe

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3 years ago

When hydrogen peroxide (H2O2) is used in rocket fuel, it produces water and oxygen (O2): Can someone help with parts B and C?

Paladinen [302]

Answer:

Explanation: A foam forms when bubbles of a gas are trapped in a liquid or solid. In this case oxygen is generated when hydrogen peroxide breaks down into oxygen and water on contact with catalase, an enzyme found in liver. Enzymes are special protein molecules that speed up chemical reactions.

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3 years ago

Write the products and balance the equation:<br><br> C4H7OH + O2 → CO2 +H2O

skad [1K]

Answer:

C4H7OH + 6O2 => 4CO2 + 4H2O

Explanation:

hope This helps :)

8 0

3 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba

Doss [256]

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

b. Van der Waals gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

$P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}$

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

i. 273.15 K and 22.414 L.

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm$

ii. 500 K and 100 cm³ (0.1 L).

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm$

i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm$

ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm$

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

5 0

3 years ago