All categories

2 years ago

What is true about radioactive isotopes of an atom?

Chemistry

2 answers:

jenyasd209 [6]2 years ago

6 0

Answer:

#2: They break down quicker than stable isotopes. - im not sure

Explanation:

stich3 [128]2 years ago

4 0

2 is the answer. They don’t break down

You might be interested in

Differentiate between positive and negative deviation ~ <img src="https://tex.z-dn.net/?f=%20%5C%5C%20" id="TexFormula1" titl

Montano1993 [528]

Answer:

positive deviation from raoults law means when the vapour pressure is higher than the solution is expected to exhibit e.g of one is the mixture of acetone and ethanol

negative deviation means that it is opposed to the total vapour pressure of the solution

7 0

2 years ago

What do all substances in the universe have in common?

Dmitry [639]

Answer:

All are made of matter

Explanation:

7 0

3 years ago

The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2

Kruka [31]

Hey there!

For this we can use the combined gas law:

$\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

We are only working with pressure and temperature so we can remove volume.

$\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}$

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

$\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}$

Solve for T₂.

$2atm = \frac{2.2atm}{T_{2}}*27C$

$2atm * T_{2}={2.2atm}*27C$

$T_{2}={2.2atm}\div2atm*27C$

$T_{2}=1.1*27C$

$T_{2}=29.7C$

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

3 0

2 years ago

The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution,

Kamila [148]

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Explanation:

Assuming the chemical equation of the compound forming product gases is:

$\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)$

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

$p_{Total}=726torr\\P_{water}=23.8torr\\$

So, pressure of nitrogen gas will be = $p_{Total}-p_{water}=726-23.8=702.2torr$

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)

V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of nitrogen gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol$

To calculate the mass of nitrogen gas, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

$0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g$

To calculate the mass percent of nitrogen gas in compound, we use the equation:

$\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100$

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%$

Hence, the mass percent of nitrogen gas in the compound is 13.3 %

8 0

3 years ago

What is the partial pressure of helium gas in a mixture that contains 8.00 grams of helium and 8.60 mol argon gas with a total p

Agata [3.3K]

Answer:

$P_{He}=9.06torr$

Explanation:

Hello there!

In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:

$P_{He}=x_{He}P_T$

Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:

$n_{He}=8.00g\frac{1mol}{4.00g}=2.00mol\\\\ x_{He}=\frac{n_{He}}{n_{He}+n_{Ar}} \\\\ x_{He}=\frac{2.00mol}{2.00mol+8.60mol}\\\\x_{He}=0.189$

Then, we calculate the partial pressure as shown below:

$P_{He}=0.189 *48.0torr\\\\P_{He}=9.06torr$

Best regards!

4 0

3 years ago