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3 years ago

8

List of things that you’ll use in the scientific method

1 answer:

MrRissso [65]3 years ago

8 0

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

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A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem

Leto [7]

Answer : The final temperature of the copper is, $87.29^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = $0.0923cal/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)$

$T_{final}=87.29^oC$

Therefore, the final temperature of the copper is, $87.29^oC$

4 0

3 years ago

1.Which is the largest atom in Group 1?

Serhud [2]

1. Francium
2. Nitrogen
3.rubidium
4. Krypton

7 0

2 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][Br^-]$

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][Br^-]$

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %

3 0

2 years ago

From what material does an extrusive rock form

Tom [10]

Lava I think try that hope it helps :)

3 0

2 years ago

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How many formula units are in<br> 35.0 g KNO3?<br> (KNO3, 101.11 g/mol)<br><br> [?]×10¹²] fun KNO3

alina1380 [7]

mol=35:101.11 g/mol=0.346

formula units = 0.346 x 6.02 x 10²³ = 2.082 x 10²³

3 0

1 year ago