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3 years ago

What subatomic particles are found in the nucleus and in the electron cloud

Chemistry

2 answers:

Dafna1 [17]3 years ago

6 0

Answer:

Electrons

Explanation:

alex41 [277]3 years ago

6 0

Someone already answered the question but I need those 10 points so hello (:

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CH4O model please ...............

zepelin [54]

Carbon hydrogen 4 Oxygen

6 0

3 years ago

Status: No credit Points earned: 0.00/1.00

DIA [1.3K]

Answer:

Cu₂O.

Explanation:

copper = 7.298

copper oxide = 8.217

oxygen in copper oxide = .919 g

7.298 g of copper = 7.298 / 63.5 mole of copper

= .1149 mole of copper

.919 g of oxygen = .919 / 16 mole of oxygen

= .0574 mole of oxygen

ratio of mole of copper and oxygen

= .1149 / .0574 = 2

formula of copper oxide = Cu₂O.

8 0

4 years ago

How many resonance structures are possible for hydroxide ion? A. 3 B. none C. 2

m_a_m_a [10]

First, how many valence electrons does hydroxide ion have? If it has something like 6 it would need 2 to get the max total of 8. Or perhaps it's 7. Then only one is needed. But if the number is 8 then 0 is required

5 0

4 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

4 years ago

Sodium acetate never precipitates True or False?

Reptile [31]

Answer:

it is false Sodium acetate NEVER precipitates

3 0

3 years ago

Read 2 more answers