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KonstantinChe [14]
3 years ago
11

Sandy puts a paper clip on the table. Then, she puts a magnet near the paperclip. The paperclip begins to move.

Chemistry
2 answers:
Akimi4 [234]3 years ago
3 0
Applied forces/or unbalanced:i hope that helps you
Marrrta [24]3 years ago
3 0

Answer: The answer is unbalanced.

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Object A has a mass of 3,600 kilograms. Object B has a mass of 600 kilograms. The weight of Object B will be ________ times the
Natalija [7]

Hi the answer is 1/6, because 1/6 of 3,600 is 600, so it's 1/6. Have a great day!


7 0
3 years ago
What is density? A) mass divided by volume B) volume divided by mass
wolverine [178]

Answer:

A) mass divided by volume

Explanation:

7 0
3 years ago
Read 2 more answers
For a reaction 2A + B 2C, with the rate equation: Rate = k[A]2[B]
In-s [12.5K]
For reaction

2 A + B  ------------> 2 C  

Rate = K [ A ]² [ B ]

<span> the order with respect to A is 2 and the order overall is 3.
</span>
hope this helps! 


3 0
2 years ago
What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
asambeis [7]
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~ </span>
6 0
3 years ago
A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o
statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = 61.1^{0}C

Heat absorbed by Scandium = 1.50*10^{3}J

The equation showing the relationship between heat, mass, specific heat and temperature change:

Q = m C (deltaT)

Where Q is heat = 1.50*10^{3}J

m is mass = 50.0 g

ΔT = 61.1^{0}C

On plugging in the values and solving for C(specific heat) we get,

1.50*10^{3}J=50.0g(C)(61.1^{0}C)

C = 0.491\frac{J}{g^{0}C }

Specific heat of the metal = 0.491\frac{J}{g^{0}C }

7 0
3 years ago
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