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insens350 [35]
2 years ago
6

Which is more likely to melt a cup of ice - a spoonful of boiling water, or a cup of room-temperature

Chemistry
1 answer:
miskamm [114]2 years ago
3 0

Answer:

boiling water

Explanation:

its hotter

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What cellular activities could thiG be a start of?
slega [8]

Could you attach a picture because I can tell you didn't post the entire question.

7 0
3 years ago
Boron has two isotopes, Boron 10 which has a mass of 10.0129 amu and Boron-11 with a mass of 11.0093 amu. B-10 occurs 84.74% of
podryga [215]

The average atomic mass of Boron: 10.431 amu

<h3>Further explanation   </h3>

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.  

So Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu  

Mass atom X = mass isotope 1 . % + mass isotope 2.%  + ...

The average atomic mass of boron :

\tt avg~mass=0.8474\times 10.0129+0.1526\times 11.0093\\\\avg~mass=10.431~amu

3 0
2 years ago
Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.
Contact [7]

Answer:

0.42 M

Explanation:

The reaction that takes place is:

  • Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:

(200 mL = 0.200L)

  • 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:

  • 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:

  • 0.224 mol - 0.14 mol = 0.085 mol

Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:

  • 0.085 mol / 0.200 L = 0.42 M

4 0
3 years ago
Determine the percentage composition (by mass) of oxygen in NH4NO3.
Brrunno [24]

Answer:

60%

Explanation:

M(NH4NO3) = 2*14 +4*1 +3*16 = 80 g/mol

M(3O) = 3*16 = 48 g/mol

(48/80) *100 % =60% oxygen by mass.

6 0
2 years ago
Ask one question you wanna know about me
gavmur [86]
Uhh how was your day or how is it going
6 0
2 years ago
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