1) Os-182 has a half-life of 21.5 hours. How many grams of a 10 g sample would have decayed after

Methylation is a process that involves which of the following metals

Vesnalui [34]

Methylation is a process that involves Mercury metal. Methylation is a process by which a methyl groups are added to the DNA molecule. when it occurs it can change the activity of a DNA segment without changing the sequence, additionally, when located in a gene promoter it may act to repress gene transcription.

7 0

4 years ago

A sample of an ideal gas has a volume of 0.500 L at 25 degrees celcius and 1.20 atm pressure. What is its volume at 75 degrees c

NemiM [27]

The ideal gas law is PV= nRT.
First you need to manipulate the equation to splice for volume,
Which will be V= nRT/P

Now you need to input the numbers for each variable. Make sure to remember what the value R equals and it’s units. R= 0.08206 L•atm/n•K

4 0

3 years ago

The universe contains only about one-tenth as many helium atoms as hydrogen atoms. But a hydrogen atom is only one-quarter the m

oksano4ka [1.4K]

Answer:

the ratio of the universe's helium to hydrogen by mass is 2:5 approximately.

Explanation:

To solve this question, we assign arbitrary values to these elements in tandem to the data given in the question.

nH, Number of Hydrogen atoms = X = 100

nHe, Number of Helium atoms = 0.1X = Y = 10

From the data given, by mass:

if the mass of Hydrogen atom = 1g = 100g

therefore, the mass of a Helium atom = 4g (1*4) = 40g

Mass ratio of Helium to Hydrogen = 40:100 = 4:10 = 2:5 approximately

8 0

3 years ago

The process by which a semipermeable membrane allows water molecules, small molecules, and ions to pass through while retaining

Jet001 [13]

The process is know as diffusion in which substances move from higher concentration to lower concentration

5 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago