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Studentka2010 [4]
3 years ago
7

Hey what's 10 & 11 y'all im dumb

Chemistry
1 answer:
lesya [120]3 years ago
3 0

Answer:

10 is 2.89g/ml

11 in attachment

Explanation:

10. D=M/V

M=250G-120G=130G

V=75ML-30ML=45ML

D=130/45=2.89g/ml

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The chemical formula for copper (II) phosphate is Cu3(PO4)2. What is the charge on each copper ion?
mrs_skeptik [129]
3+ because I just know it 8757890
6 0
3 years ago
What minimum energy is required to excite a vibration in HF?
Elodia [21]

Answer:

The energy of a vibrating molecule is quantized much like the energy of an electron in the hydrogen atom. The energy levels of a vibrating molecule are given by the equation: En=(n+21)hv where n is a quantum number with possible values of 1, 2, ... and v is the frequency of vibration.

Explanation:

hope it helps.

have a wonderful day!

7 0
3 years ago
The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


3 0
3 years ago
I. Give the name for the following compounds:
alexandr402 [8]

Answer:

Explanation:

1) MgBr₂

Magnesium bromide

2) Ca₃(PO₄)₂

Calcium phosphate

3) NO₂

Nitrogen dioxide

4) Ni₂(CO₃)₃

Nickel(III) carbonate

5) (NH₄)₂CO₃

Ammonium carbonate

6) MnBr₂

Manganese bromide

7) Na[MnO₄]

Sodium permanganate

8) P₂O₅

Phosphorus pentoxide

9) CCl₄

Carbon tetrachloride

10) Fe(OH)₂

Iron(ii) hydroxide

8 0
3 years ago
At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2
xxMikexx [17]

Answer:

K = 3.37

Explanation:

2 NH₃(g) → N₂(g)  + 3H₂(g)

Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).

              2 NH₃(g)    →    N₂(g)  + 3H₂(g)

Initally       4moles             -            -

React        2moles           2m   +   3m

Eq             2 moles          2m        3m

We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)

The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²

We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)

K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²

K = 27/8 / 1 → 3.37

5 0
3 years ago
Read 2 more answers
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