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4 years ago

Please help I need a 100!!!

Chemistry

2 answers:

Salsk061 [2.6K]4 years ago

6 0

Its in vertical columns. I hope this helps and you get a 100 on your test!

Artemon [7]4 years ago

5 0

C) in vertical columns

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Suppose that a certain fortunate person has a net worth of $79.0 billion ($7.90×1010). If her stock has a good year and gains $3

mario62 [17]

A net worth: $79.0 billion.
Value of stock : $3.20 billion.
New net worth:
$79.0 + $3.20 = $82.20 billion = $82.20 * 10^9 = $8.20 * 10^10

7 0

4 years ago

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0

3 years ago

We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim

eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know

Ku = Vp * Kair + Vmagnesium * K metal

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0

3 years ago

Calculate the concentration in % (m/m) of a solution containing 30.0g of mgcl2 dissolved in 270.0g of h20

qaws [65]

As the question tells you, you need to use the formula

% mass= mass of solute/ mass of solution x 100

mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g

% mass= 30.0/ 300.0 x 100= 10%

answer is B

4 0

3 years ago

What is the concentration of the murexide soution with a transmittance of 28.65%. (molar absorptivity = 3847)

Ainat [17]

Answer:

The concentration of the murexide solution is 0.0000745 M

Explanation:

From Beer-Lambert's law,

A = εlc

A = Absorbance = 28.65% = 0.2865

ε = molar absorptivity = 3847 M/cm

l = path length = 1cm

c = concentration in mol/L = ?

c = A/εl = 0.2865/(3847×1) = 0.0000745 mol/L

Hope this Helps!

6 0

3 years ago