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3 years ago

What is the volume 0.0023 moles of CO2

1 answer:

4 0

1 mole ---------------- 22.4 ( at STP )
0.0023 moles -------- ?

v = 0.0023 * 22.4 / 1

v = 0.05152 L

hope this helps!

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A 100 g sample of an unknown liquid absorbs 2000 j of heat energy, raising the liquid's temperature from 50 ◦ c to 70 ◦

Mekhanik [1.2K]

Since there is no phase change, we can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J kg⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Q = 2000 J
m = 100 g = 0.1 kg
c = ?
ΔT = (70 °C - 50 °C) = 20 °C

By applying the formula,
2000 J = 0.1 kg x c x 20 °C
c = 2000 J / (0.1 kg x 20 °C)
c = 1000 J kg⁻¹ °C⁻¹

Hence, the specific heat capacity of the liquid is 1000 J kg⁻¹ °C⁻¹.

5 0

4 years ago

The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5

Deffense [45]

Given: Half-life of 133Ba = t1/2 = 10.5 years.

The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.

Also, $k = \frac{2.303}{t} log \frac{Co}{Ct}$
where, Co = initial concentration = 1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................

Answer 1: For t = 5 years
 $0.066 = \frac{2.303}{5} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.

Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................

Answer 2: For t = 30 years
$0.066 = \frac{2.303}{30} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.

Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................

Answer 3: t = 180 years
$0.066 = \frac{2.303}{180} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.

Number of 133Ba atoms left after 180 years = 7.6367 X10^4.

8 0

4 years ago

How many moles of naoh are present in 29.0 ml of 0.290 m naoh?

Dovator [93]

29.0 mL in liters:

29.0 / 1000 => 0.029 L

n = M x V

n = 0.290 x 0.029

n = 0.00841 moles of NaOH

hope this helps!

7 0

3 years ago

Calculate the percent ionization of benzoic acid at the following concentrations. (a) 0.32 M WebAssign will check your answer fo

seraphim [82]

Answer: a) 1.4 %

b) 48%

Explanation:

$C_7H_6O_2\rightarrow H^+C_7H_5O_2^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

a) Given c= 0.32M and $K_a=6.3\times 10^{-5}$

$\alpha$ = ?

Putting in the values we get:

$6.3\times 10^{-5}=\frac{(0.32\times \alpha)^2}{(0.32-0.32\times \alpha)}$

$(\alpha)=0.014$

$\%(\alpha)=0.014\times 100=1.4\%$

b) Given c= 0.00014 M and $K_a=6.3\times 10^{-5}$

$\alpha$ = ?

Putting in the values we get:

$6.3\times 10^{-5}=\frac{(0.00014\times \alpha)^2}{(0.00014-0.00014\times \alpha)}$

$(\alpha)=0.48$

$\%(\alpha)=0.48\times 100=48\%$

8 0

3 years ago

Read the sentence.

ExtremeBDS [4]

Answer:

The correct answer is "reveal the structures".

Explanation:

The predicate is the part of the prayer that tells us what the subject is doing. In this prayer "The Fossils" is the subject, so we must seek the action that develops from the subject. That action is that the fossils "reveal the structures". In this way, we can find the simple predicate of this sentence.

Have a nice day!

3 0

3 years ago