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hjlf
2 years ago
5

Help!!!

Chemistry
1 answer:
TEA [102]2 years ago
4 0
Is b decrease because it is right
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If a uranium atom undergoes both alpha and gamma decay, what happens to it?
WARRIOR [948]

If a uranium atom undergoes both alpha and gamma decay, then it means that there will be formation of one helium particle which is also known as alpha particle and gamma decay is the radiation or release of energy. Whereas in a radioactive reaction, Uranium-235 absorbs a neutron and splits into two new atoms.

hope this helps

5 0
3 years ago
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COCI2 has an effusion rate of 0.00172 m/sec. Which of the gases below would have an effusion rate of 0.00323 m/sec?
geniusboy [140]
CO por qué si y punto, chao
8 0
2 years ago
Can you please help me and can you show your work please
natali 33 [55]

Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

4. 340.6 grams of ammonia.

5. 4.5x10²³ molecules of sulfur dioxide.

Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

Best regards!

5 0
3 years ago
Which element or ion listed below has the electron configuration 1s22s22p6?
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Count up the number of electrons in each orbital. theres ten electrons which means it has 10 protons. the element would be neon
7 0
3 years ago
A gas occupies a 1.5 L container at 25 degrees Celsius and 2.0 atm. If the gas is transferred to a 3.0 L container at the same t
harkovskaia [24]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1 x V1 / V2

P2 = 2.0 x 1.5 / 3

<span>P2 = 1 atm</span>

4 0
3 years ago
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