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cricket20 [7]
3 years ago
13

Why are some metals such as copper and aluminium not magnetic​

Chemistry
1 answer:
ioda3 years ago
8 0

Answer:i dont know

Explanation:

I D

K

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List of things that you’ll use in the scientific method
MrRissso [65]

Explanation:

Make an observation

ask a question

form a hyposisese

make a prediction

test the prediction

use the results to fix the prediction and hypotheses

8 0
3 years ago
Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures.N2(g) + O2(g) 2NO(g)The
yanalaym [24]

Answer : The correct option is, (E) 7.8 atm

Explanation :

The partial pressure of N_2 = 8.00 atm

The partial pressure of O_2 = 5.00 atm

K_p = 0.0025

The balanced equilibrium reaction is,

                               N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initial pressure     8.00      5.00            0

At eqm.               (8.00-x) (5.00-x)        2x

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}

Now put all the values in this expression, we get :

0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}

By solving the terms, we get:

x=0.15atm

The equilibrium partial pressure of N_2 = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of N_2 is 7.8 atm.

3 0
4 years ago
Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h
Alborosie
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
\frac{number of moles of [OH-]}{volume of solution (L)} = \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} = 0.03199 M
6 0
3 years ago
If 4.70 L of CO2 gas at 22 ∘C at 789 mmHg is used, what is the final volume, in liters, of the gas at 37 ∘C and a pressure of 75
lutik1710 [3]

Answer:

Final Volume = 5.18 Liters

Explanation:

Initial Condition:

P1 = 789 mm Hg x (1/760) atm /mm Hg = 1.038 atm

T1 = 22° C = 273 + 22 = 295 K

V1 = 4.7 L

Final Condition:

P2 = 755 mm Hg x (1/760) atm /mm Hg = 0.99 atm

T2 = 37° C = 273 + 37 = 310 K

V2 = ?

Since, (P1 x V1) / T1 =  (P2 x V2) / T2,

Therefore,

⇒ (1.038)(4.7) / 295 = (0.99)(V2) / 310

⇒ V2 = 5.18 L (Final Volume)

5 0
3 years ago
An unknown solution has a pH of 8. How would you classify this solution?
ratelena [41]
The correct answer is B) Basic. Hope this helps.
6 0
3 years ago
Read 2 more answers
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