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3 years ago

How do u release a genjutsu when an comrade is caught in it?

Chemistry

1 answer:

NNADVOKAT [17]3 years ago

7 0

Your best answer is choice D.

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Explain how the removal of thermal energy (heat) from matter affects its particles

den301095 [7]

Answer:

You can change an object's state of matter by adding or removing thermal energy. When you add thermal energy to an object, these things can happen: Particles move faster

Explanation:

6 0

2 years ago

Plz help me plz I am timed!!

PolarNik [594]

Answer:

full moon

Explanation:

....ur welcome..............

6 0

2 years ago

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Charlotte is driving at 58.6 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 4

kramer

Answer: 19.71 feet

Explanation:

Given: Speed of Charlotte =58.6 mi/h

Since 1 hour = 3600 seconds

and 1 mile = 5280 feet

So, Speed of Charlotte = $\dfrac{58.6\times5280}{3600}\approx85.95$ mi/ sec

She looks down at her phone and takes her eyes off the road for 4.36 s.

Since , Distance = $\dfrac{speed}{time}$

So, Distance = $\dfrac{85.95}{4.36}=19.71\ feet$

Hence, Charlotte traveled 19.71 feet during this time.

8 0

2 years ago

What is considered to be a physical property of matter

katrin [286]

Answer: Some examples are color, density, volume and mass

Explanation:

Physical properties are anything you can smell, touch, or hear. They can be observed without changing.

8 0

3 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

3 years ago

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