All categories

3 years ago

PLEASE!!! I NEED HELP ASAP!!

Chemistry

2 answers:

Lana71 [14]3 years ago

7 0

W=m₁/m₀=2^(-t/T)

t=4.6·10⁹ years
T=5·10¹⁰ years

w=2^(-4.6·10⁹/5·10¹⁰)

w=0.9382
w=93.82%

iris [78.8K]3 years ago

5 0

Answer:

the ratio of parent to daughter = 15.38 : 1

Explanation:

The radioactive disintegration is considered to be first order reaction.

The relation between first order rate constant and the half life is:

$k=\frac{0.693}{t_{\frac{1}{2} }}$

Let us calculate the rate constant from half life as

$k=\frac{0.693}{50}=0.01386 (billionyr)^{-1}$

The rate law expression for first order reaction is

$kt=ln\frac{[A0]}{At}$

Where

A0 = initial amount of rubidium isotope = say 100

At= final amount of rubidium isotope after 4.6 billion years

putting all values

$0.1386X4.6=ln\frac{100}{At}$

1.065=100/At

At=93.897

The amount of daughter isotope formed =100-93.897 = 6.103

the ratio of parent to daughter = 93.897:6.103 = 15.38 : 1

You might be interested in

Nuclei with the same mass number but different atomic numbers are called isobars. Consider Ca-40, Ca-41, K-41 and Ar-41. (a) Whi

SVETLANKA909090 [29]

Answer:

Part a:

Isobars: Ca-41, K-41 and Ar-41

Part b:

Number of proton

Part c:

Incorrect statement

Explanation:

Part a:

Nuclei with same mass number are called isobars.

Given:

Ca-40, Ca-41, K-41 and Ar-41

Ca-41, K-41 and Ar-41 have equal mass numbers, so these are isobars.

Part b:

Atomic number of Ca = 20

Atomic no. = Number of proton.

So, Ca-40 ans C-41 have same number of proton or in other words proton count is common in both.

Part c:

Number of neutrons in Ca-41

Atomic number = 20

Mass number = 41

Number of neutron = Mass number -atomic number

= 41 -20 = 21

Number of neutrons in K-41

Atomic number = 19

Mass number = 41

Number of neutron = Mass number -atomic number

= 41 -19 = 22

Number of neutrons in Ar-41

Atomic number = 18

Mass number = 41

Number of neutron = Mass number -atomic number

= 41 -18 = 23

So, the statement is incorrect.

3 0

3 years ago

Help science worth 50 points

monitta

Answer:

I think this is like a game in which you need dice.

Explanation:

To fill out the second page, you need to complete the roll/go-to chart.

8 0

4 years ago

Read 2 more answers

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago

How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th

OLEGan [10]

Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>

<em />

We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

$48.3g.\frac{1mol}{46.07g} =1.05mol$

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

3 0

4 years ago

shusha [124]

Answer:

Salt Mass: 58.44 g/mol Melting point: 801 °C

Explanation:

5 0

4 years ago

Read 2 more answers