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Dmitry_Shevchenko [17]
3 years ago
8

1. Write the molecular balanced equation for the dissolution of copper. 2. Write the complete ionic balanced equation for the di

ssolution of copper. 3. Write the net ionic balanced equation for the dissolution of copper. 4. What type of reaction is this
Chemistry
1 answer:
olganol [36]3 years ago
8 0

Answer:

Cu + HNO3 -----> Cu (NO3) 2 + NO2 + H2O

Explanation:

This equation is the equation for dissolving copper in nitric acid, this reaction is a REDOX reaction, that is, an oxide-reduction reaction.

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Help me with this please!!! Smart ones help me 18 points
Helen [10]
Hey! The answer would be 1/2.
4 0
3 years ago
Read 2 more answers
Calculate the density of an aluminum block with a volume of 100 cm3 and a mass of 270g.
m_a_m_a [10]

Answer:

<h2>2.7 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{270}{100}  =  \frac{27}{10} \\

We have the final answer as

<h3>2.7 g/cm³</h3>

Hope this helps you

6 0
2 years ago
The chemical formula for disulfur monoxide is:<br> A. S2O<br> B. 2SO<br> C. SO2<br> D. S2O2
Delvig [45]
The answer is (A).
Hope this helps :).

3 0
3 years ago
When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the ph
Scrat [10]

Answer:

E = 12.92 × 10^(-16) J

Explanation:

Formula for energy is;

E = hc/λ

Where;

h is Planck's constant = 6.63 x 10^(-34) J.s

c is speed of light = 3 × 10^(8) m/s

λ is wavelength = 0.154 nm = 0.154 × 10^(-9) m

E = (6.63 x 10^(-34) × 3 × 10^(8))/(0.154 × 10^(-9))

E = 12.92 × 10^(-16) J

6 0
3 years ago
If carbon-15 has a half-life of 2.5 seconds, % of the sample would still be
kramer

Answer:

After 5 second 25% C-15 will remain.

Explanation:

Given data:

Half life of C-15 = 2.5 sec

Original amount = 100%

Sample remain after 5 sec = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 5 sec / 2.5 sec

Number of half lives = 2

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

Thus after 5 second 25% C-15 will remain.

8 0
3 years ago
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