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3 years ago

PLEASE HELP!!!

Chemistry

1 answer:

Elena L [17]3 years ago

3 0

Answer:

no it should not that place is historical and so they should make where u can visit but protect it as a historical landmark

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1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.

maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.

Explanation:

The molecular formula of boron trifluoride is $BF_3$ .

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- $6.023x10^2^3$ atoms.

Then, 2.20 moles of boron has

- $=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms$

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

$(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol$

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0

3 years ago

Someone help me plz :))) brainliest

SOVA2 [1]

Answer:

Option 2

Explanation:

3 0

3 years ago

A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?

jeyben [28]

STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c. so we have,

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P

substitute the values into the equation,
c = [(0.63g/L)(0.08206 L-atm/mol-K)(273K)]/(1atm)
c = 14.11 g/mol

3 0

3 years ago

1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a

Archy [21]

Answer:

Explanation:

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃ + Al → 3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

Al : Ag

1 : 3

0.223 : 3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂ → 2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

HgO : Hg

2 : 2

3.12 : 3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂ + 5O₂ → 2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

N₂O₅ : O₂

2 : 5

12.99 : 5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

C₆H₆ : CO₂

2 : 12

0.103 : 12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0

3 years ago

Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented

Yuki888 [10]

17.8 mL NaOH

Step 1. Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

Step 3. Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

Step 4. Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

3 0

3 years ago