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3 years ago

If you start with 42 g of Fe-53, how much is left after 8.51 minutes? I

Chemistry

2 answers:

lidiya [134]3 years ago

5 0

Answer:

2.58 good luck with everything

Paladinen [302]3 years ago

4 0

Answer:

2:38

Explanation:

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When a candle burns it produced 41,300 Joules per 1 gram. Use dimensional analysis to convert this to Calories per pound.

insens350 [35]

Answer:

4477381.7 calories/pound

Explanation:

It is given that,

When a candle burns it produced 41,300 Joules per 1 gram.

We need to convert it into calories per pound.

We know that,

1 cal = 4.184 J

⇒ 1 J = (1/4.184) cal

1 pound = 453.592 grams

⇒1 g = (1/453.592) pounds

Now,

$41300\ \text{Joules per gram}=\dfrac{41300\ \text{Joules}}{\text{gram}}\\\\=41300\times \dfrac{\dfrac{1}{4.184}\ \text{calories}}{\dfrac{1}{453.592}\ \text{gram}}\\\\=4477381.7\ \text{calories/pound}$

Hence, 41,300 Joules/gram = 4477381.7 calories/pound.

3 0

3 years ago

A compressed gas cylinder is filled with 5270 g of argon gas. The pressure inside the cylinder is 2050 psi at a temperature of 1

LuckyWell [14K]

Answer:

A compressed gas cylinder is filled with 5270 g of argon gas.

The pressure inside the cylinder is 2050 psi at a temperature of 18C.

The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.

How many grams of argon remains in the cylinder?

Explanation:

First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:

Mass of Ar gas is --- 5270g.

The number of moles of Ar gas:

$Number of moles of Ar gas=\frac{given mass of Ar}{its atomic mass} \\ =\frac{5270g}{39.948g/mol} \\ =131.9mol$

Temperature T=(18+273)K=291K

Pressure P=2050psi

$2050* 0.0680atm\\\\=139.4atm\\$

Volume V=?

$PV=nRT\\139.4atm * V=131.9mol *0.0821L.atm.mol-1.K-1 * 291K\\=>V=\frac{131.9mol *0.0821L.atm.mol-1.K-1 * 291K}{139.4atm} \\=>V=22.6L$

Using this volume V=22.6L

Pressure=650psi=44.2atm

Temperature T= (26+273)K=299K

calculate number of moles "n" value:

$PV=nRT\\=>n=\frac{PV}{RT} \\=>n=\frac{44.2atm*22.6L}{0.0821L.atm.mol^-1K^-1* 299K} \\=>n=40.7mol$

Mass of 40.7mol of Ar gas:

$mass of Ar gas=number of moles * its atomic mass\\\\ =40.7mol* 39.948g/mol\\\\ =1625.8g$

Answer:

The mass of Ar gas becomes 1625.8g.

8 0

2 years ago

Density of 2.7 g/ml and volume of 35.6 ml what is the mass

Anettt [7]

The math is set up like

35.6 ml * 2.7 g/ 1 ml

which will leave you with

96.12 g

3 0

3 years ago

3. A sample of carbon has a mass of

Sav [38]

$- 34 {}^{15}$

The power of -34 will be 15 because 15 zeroes are given....

8 0

2 years ago

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

2 years ago

If you start with 42 g of Fe-53, how much is left after 8.51 minutes? I​

If you start with 42 g of Fe-53, how much is left after 8.51 minutes? I