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mario62 [17]
2 years ago
14

3.50 grams of an unknown metal was placed in a graduate cylinder containing 50.0

Chemistry
1 answer:
Olenka [21]2 years ago
3 0

For 3.50 grams of an unknown metal was placed in a graduate cylinder containing 50.0mL water initially,  the density of this unknown meta  is mathematically given as

D= 0.23 g/ml

<h3>What is the density of this unknown metal if the final reading from the graduate cylinder was 65.0 mL?</h3>

Generally, the equation for the Volume of water  is mathematically given as

Vw= Volume of water after metal addition - Volume of water initial

Therefore

Vw = 65 ml - 50 ml

Vw= 15 ml

In conclusion,

D= Mass of metal / Volume

D= 3.5 / 15

D= 0.23 g/ml

Read more about mass

brainly.com/question/15959704

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Categorize each hydrocarbon as being saturated or unsaturated.
Black_prince [1.1K]

As you have not provided the options, still we can figure out the answer by understanding the key difference between saturated and unsaturated hydrocarbons.

SATURATED HYDROCARBONS are those hydrocarbons which only consist of a carbon carbon single bonds. All the bonds are sigma there are no pi bonds at all. Examples are shown below.

While, UNSATURATED HYDROCARBONS are those hydrocarbons which may contain either a double bond or triple bonds or both of them between the carbon atoms as shown below.

6 0
3 years ago
If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?
natulia [17]
Volume = a x a x a  

V = 2 cm x 3 cm x 4 cm => 24 cm³

Density = 19.3 g/cm³

Mass = ?

Therefore:

m = D x V

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4 0
3 years ago
How do you view failure? Is failure the end, or is failure an important lesson and motivator?
Katyanochek1 [597]

Answer:

Failure is a lesson

Explanation:

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8 0
3 years ago
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

5 0
3 years ago
GIVE ME ANSWERS
Shalnov [3]

Answer:

uh.. How many do you want done??

Explanation:

6 0
3 years ago
Read 2 more answers
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