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3 years ago

Forces that change an object motion by touching the object are _______ forces

1 answer:

7 0

Forces are pushes and pulls that may change the motion of an object. Balanced forces result in an object remaining at rest or moving at a constant speed. Unbalanced forces result in the acceleration of an object. An object's motion depends on how it changes position.

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In an investigation, explain the best way to use variables in an experiment

vazorg [7]

Answer:

An experiment is an investigation in which a hypothesis is scientifically tested. In an experiment, an independent variable (the cause) is manipulated and the dependent variable (the effect) is measured; any extraneous variables are controlled. An advantage is that experiments should be objective.

Explanation:

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2 years ago

After some wood has completely burned in a fire some of the matter that made up the wood has been destroyed

STatiana [176]

Answer:

Because it went through a chemical change which changes its atomic form

5 0

2 years ago

C2h4 3 o2 2 co2 2 h2o

Marat540 [252]

Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2

O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving

0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2

3 0

3 years ago

4) The initial rate of the reaction between substances P and Q was measured in a series of

ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

$rate = k[P]^{2} [Q]$

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

$rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }$

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

$k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4$

Second row:

2. Rate value:

$rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1$

3.Third row:

$[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}$

4. Fourth row:

$[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}$

6 0

3 years ago

Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

$Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}$

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

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3 years ago