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larisa86 [58]
3 years ago
9

Indicar que elemento es de transición *

Chemistry
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

c hierro

no se la otra sorry

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A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi
Naddika [18.5K]
Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span>molality =n(solute) / m(solvent) =\ \textgreater \  4.8m = 4.8mole / 1kg
= 4.8 mole * 98 g/mole = 470g
m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \=\ \textgreater \  Molarity = 4.8 mole / 1.2 L = 4 M m(H2SO4)  which is =<span>470g

</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
Molarity = n(solute) / V(solution) =\ \textgreater \
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3 years ago
How are the orbits of asteroids similar to, or different from planetary orbits?
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3 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
4 years ago
What is the correct formula for sodium carbonate?
frosja888 [35]

Answer:

na2 co3

Explanation:

sodium carbonate formula

4 0
3 years ago
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A lake has been infected by some type of new algae that is unknown. Every single day the amount of surface area that the algae t
FrozenT [24]

Answer:

It takes 86 days take to cover half of the lake

Explanation:

In the day #1, the amount of the algae is X,

In the day #2 is 2X

In the day #3 is 2*2*X = X*2²

...

In the day #n the amount of the algae is X*2^(n-1)

Assuming X = 1m³. In the day 87, the area infected was:

1m³*2^(87-1)

7.74x10²⁵m³ is the total area of the lake

the half of this amount is 3.87x10²⁵m³

The time transcurred is:

3.87x10²⁵m³ = 1m³*2^(n-1)

Multiplying for 5 in each side:

ln (3.87x10²⁵) = ln (2^(n-1))

58.9175 = n-1 * 0.6931

85 = n-1

86 = n

<h3>It takes 86 days take to cover half of the lake</h3>

4 0
3 years ago
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