2NaOH + H2So4=_____

How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102

Ede4ka [16]

Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles$

1 mole of hydrogen $(H_2)$ = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms

17.5 mole of hydrogen $(H_2)$ = $\frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25}$ atoms

There are $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0

3 years ago

1. identify whether following have ionic or covalent bonding. Give your reasons: Oxygen molecule, water, calcium oxide, hydrogen

Romashka [77]

First you need to know the different between an ionic and covalent bond. An ionic bond is the pairing of a metal and non-metal element. A covalent bond is the pairing of 2 nonmetals.

Metals are the elements at the left of the periodic table while non-metals are the elements at the right of the periodic table.

You should also know the diatomic (di means 2) molecules also known as the fab 7. These molecules will always form covalent bonds. These molecules are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. With the subscripts, these molecules would be written as H ₂, N ₂, O ₂, F ₂, Cl ₂, Br ₂, and I ₂.

5 0

3 years ago

The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.

Assoli18 [71]

a) To find the mass after t years:

we will use this formula:

A = Ao / 2^n

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)

b) Mass after 90 y :

by using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg by solving for t

∴ t = 229 y

7 0

3 years ago

1.25 cm is the same distance as ?<br> 12.5 Km<br> 12.5 mm<br> 12.5 dm<br> 12.5 m

pogonyaev

Answer:

12.5mm

Explanation:

1cm = 10mm

so you need to multiply the 1.25 by 10 to get it in mm.

6 0

3 years ago

The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t

Katena32 [7]

The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.

<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

655/12 x 6 + 12+ 16 x 6
= 655/ 180 = 3.6888 moles.

2NaOH + H2So4=______+2H2O​

2NaOH + H2So4=______+2H2O