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EastWind [94]
3 years ago
13

A rock displaces 1,500 mL of water. The volume of the rock is

Chemistry
2 answers:
Tatiana [17]3 years ago
5 0
The answer is:  [A]:  1,500 cm<span>³ .
_______________________________________
Explanation:
_______________________________________
1 mL = 1 cm</span><span>³ ; 

So 1,500 mL = 1,500 cm</span><span>³ .
_______________________________________</span>
Schach [20]3 years ago
3 0
1500 cm^3 ; 1 mL equals 1 cm^3
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The first eighteen chemical elements on the periodic table are described below in a series of statements. The first 1S letters o
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Answer: The questions looks unclear

Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.

1. D belongs to group 4

E. Belongs to group 7

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A belongs to group 8. A noble gas.

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3 0
3 years ago
A volume measured by a graduated cylinder that was marked in 100 mL
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The volume measured using such a cylinder will be reported to the nearest 10th mL.

<h3>Cylinder graduation</h3>

10 mL graduated cylinders are always read to the nearest two decimal places.

100 mL graduated cylinders are always read to the nearest 1 decimal place. The nearest 1 decimal place is the same thing as the nearest 10th.

Thus, a reading made using a 100mL increment graduated cylinder would be reported to the nearest 10th mL.

More on cylinder graduation can be found here: brainly.com/question/14427988

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8 0
2 years ago
Type the correct answer in the box. Express the answer to two significant figures.
meriva

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8 0
3 years ago
Read 2 more answers
1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

___________________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

3 0
3 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
3 years ago
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