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2 years ago

What resources are sometimes shared by squirrels and certain birds answer quick I need to get to the bus stop!!

Chemistry

1 answer:

Colt1911 [192]2 years ago

7 0

Answer:

I know someone that has the answer

Explanation:

I know someone that has the answer

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When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°C to 115°C. Find the specific hea

lesantik [10]

<h3>Answer:</h3>

0.89 J/g°C

<h3>Explanation:</h3>

Concept tested: Quantity of heat

We are given;

Mass of the aluminium sample is 120 g
Quantity of heat absorbed by aluminium sample is 9612 g
Change in temperature, ΔT = 115°C - 25°C

= 90°C

We are required to calculate the specific heat capacity;

We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.

That is;

Q = m × c × ΔT

Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.

Specific heat capacity, c = Q ÷ mΔT

= 9612 J ÷ (120 g × 90°C)

= 0.89 J/g°C

Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C

4 0

3 years ago

Can someone please help me with this?

yaroslaw [1]

Answer:

Lonic.an electron will be transferred from potassium to the chlorine atom

3 0

3 years ago

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If each of you pulls with equal force, the rope will not move (the two forces are equal in strength, but opposite in direction).

Vedmedyk [2.9K]

Answer:

i dont know

Explanation:

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3 years ago

What volume will 0.875 moles of Kr occupy at STP

umka21 [38]

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3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago