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Shalnov [3]
2 years ago
13

Which piece of lab equipment listed below would be the best for measuring the volume of a liquid.

Chemistry
1 answer:
notka56 [123]2 years ago
4 0
The answer is a graduated cylinder
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Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing tin
mojhsa [17]
  The  half  reactions    as  they  occur   at  each  electrode

is  as  follows

at  the  anode  Sn(s)  =sn^2+(aq)  +  2e  -
at  the  cathode  2  ag^+(aq)   + 2e -  =  2Ag (s)

net  cell  reaction  =   Sn (s)  +  2Ag^+(aq)  =  sn^2+ (aq) +  2  Ag  (s)
8 0
3 years ago
Which particle changes the mass of the isotope the most? which particle changes the mass of the isotope the most? electron captu
MAXImum [283]
This question is absurd
7 0
3 years ago
Read 2 more answers
What is the balanced equation and type of reaction for: <br> Copper metal+aqueous silver nitrate
Gnesinka [82]

Answer:

Cu(s) + 2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

This chemical equation means:

One mole of solid copper plus two moles of aqueous silver nitrate produce one mole of copper(II) nitrate plus two moles of solid silver.

This is a single replacement reaction in which the metal copper replaces the metal silver.

3 0
2 years ago
Given the following equation: Cu + 2 AgNO3 ---&gt; Cu(NO3)2 + 2
Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

  • Molar mass of AgNO3 = 169.87 g/mol
  • Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0
2 years ago
Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>

4 0
3 years ago
Read 2 more answers
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