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2 years ago

Is there a way to stop radioactive decay?

1 answer:

4 0

The chain reaction is easy to stop. Just add a neuron absorbing material. The Control Rods in rectors can do that You just SCRAM (put the rods all the way in) or add something like Boron and the chain reaction stops.

The problem is the radioactive waste. Those isotopes break down and release heat spontaneously, no neutrons required. The only known way to stop or slow radioactive decay down is to slow time down by moving at relativistic speed or near orbit to a black hole.

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Explain what an induced current is. please help me with at least 2 sentences.

Deffense [45]

An induced current is a conducting loop that is exposed to a changing magnetic field. And it is knows as induced current

6 0

2 years ago

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A student observed tiny dots moving about when exploring a drop of pond water. At 400x magnification, the student recognized the

igomit [66]

Paramecia are examples of unicellular organisms. They are protists.

<h3>Protists</h3>

Protists are eukaryotic unicellular (single-celled) organisms that do not fit into any other category.

Paramecia are protists microorganisms generally found in aquatic habitats (e.g., a pond).

Paramecia generally have an oblong shape and they are covered with tiny structures known as cilia (for motion).

Learn more about protists here:

brainly.com/question/12206236

5 0

2 years ago

a sample of ideal gas at room temperature occupies a volume of 12.0 L at a pressure of 392 torr. if the pressure changes to 1960

Setler79 [48]

852 x 40.0= 4260 x v2
v2= 8 L 

b) 852 x 40 = p2 x 69.0 
p2= 494 torr

3 0

2 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

2 years ago

If you forget your units , is disaster guaranteed

Paha777 [63]

Yep. And you'll always get equations with units that don't match

4 0

2 years ago