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Marizza181 [45]
3 years ago
5

Bromine has an atomic number of 35. How many protons are in an atom of bromine? 17 18 35 52

Chemistry
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

There are 35 protons in a single atom of Bromine

Explanation:

The atomic number of an element represents the number of protons contained in the nucleus of the atom! Thus, since Bromine's atomic number is 35, that means there are 35 protons!

Hope this helps! :)

You might be interested in
How many molecules are there in 560 grams of CoCl2?​
Marrrta [24]

Answer:

25.89  × 10²³ molecules

Explanation:

Given data:

Mass of CoCl₂ = 560 g

Number of molecules present = ?

Solution:

Number of moles of CoCl₂:

Number of moles = mass/molar mass

Number of moles = 560 g/ 129.84 g/mol

Number of moles = 4.3 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

4.3 mol ×  6.022 × 10²³ molecules /1 mol

25.89  × 10²³ molecules

5 0
2 years ago
4. Chemical formula, mg (Cl0₃)2
grin007 [14]

Answer:

0.017mole

0.0033M

Explanation:

Given parameters:

Formula of the compound:

           Mg(ClO₃)₂

Mass of the sample  = 3.24g

Unknown:

Number of moles of the sample = ?

Molarity  = ?

Solution:

The number of moles of any substance is given as:

   Number of moles  = \frac{mass}{volume}  

 Molar mass of Mg(ClO₃)₂  = 24 + 2[35.5 + 3(16)]  = 191g/mol

    Number of moles  = \frac{3.24}{191}   = 0.017mole

Molarity is the number of moles of a solute in a solution:

    Molarity  = \frac{number of moles }{volume}  

 Volume given  = 5.08L

  Molarity  = \frac{0.017}{5.08}  = 0.0033M

5 0
3 years ago
What can be absorbed or produced as the result of a chemical reaction???
tia_tia [17]
Heat can be absorbed or produced
6 0
3 years ago
0.32 g of a walnut is burned under an aluminum can filled with 58.1 mL of water. The water temperature in the can increases by 3
balu736 [363]

Answer:

1.8 × 10² cal

Explanation:

When 0.32 g of a walnut is burned, the heat released is absorbed by water and used to raise its temperature. We can calculate this heat (Q) using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity of water

m: mass of water

ΔT: change in the temperature

Considering the density of water is 1 g/mL, 58.1 mL = 58.1 g.

Q = c × m × ΔT

Q = (1 cal/g.°C) × 58.1 g × 3.1°C

Q = 1.8 × 10² cal

3 0
3 years ago
Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: K_{c} = 2.10^{-2}

Explanation: The reaction for steam reforming methane is:

CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

4 0
3 years ago
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