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3 years ago

Which learning theory is based on associative learning?

1 answer:

4 0

Classical conditioning

Explanation:

Its a type of associative learning based on the association between a neutral stimulus with another that is significant for a person or an animal in order to generate a similar response.

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Hat is the gram formula mass of caso4 •2h2o

Trava [24]

Your answer would be 172.1703 g/ml

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3 years ago

An organism with no natural enemies is called

Archy [21]

Answer:

it is pathogens i think its that

Explanation:

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3 years ago

Read 2 more answers

Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0

nexus9112 [7]

Answer: The correct answer is Option b.

Explanation:

To calculate the amount of heat absorbed or released, we use the following equation:

$q=mc\Delta T$ .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of water = 4.186 J/g ° C

$\Delta T$ = Change in temperature

For Trial 1:

We are given:

$m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J$

Putting values in equation 1, we get:

$q=30g\times 4.186J/g^oC\times 40^oC$

q = 5023.2 J

For trial 2:

We are given:

$m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J$

Putting values in equation 1, we get:

$q=40g\times 4.186J/g^oC\times 10^oC$

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = $(5023.2-1674.4)J=3347J$

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0

3 years ago

What the first questions answer? Anyone know if so plz help a siesta out

Marysya12 [62]

Answer:

Explanation:

8 0

3 years ago

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

6 0

3 years ago