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Anestetic [448]
3 years ago
12

Which of the following is true for an atom which has 29 protons, 34 neutrons and 29 electrons?

Chemistry
1 answer:
xeze [42]3 years ago
6 0
The correct description that holds true for the above question would be B. Atomic number - 29 and Mass Number - 63.
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Convert 1.75 g/mL to the unit g/m3. Answer Choices: (1.75 x 106 g/m3), (1.75 x 102 g/m3), (1.75 x 10-2 g/m3), (1.75 x 10-6 g/m3)
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4 years ago
A sample of gas contains 0.1100 mol of HBr(g) and 5.00x10^-2 mol of Cl2(g) and occupies a volume of 5.85 L. The following reacti
Vesna [10]

The volume of the sample after the reaction : V₂=5.484 L

<h3>Further explanation</h3>

Given

Reaction

2HBr(g) + Cl2(g) => 2HCl (g)+Br2(g)

0.1100 mol of HBr(g)

5x10⁻² mol of Cl2(g)

Required

The volume of the sample after reaction

Solution

Find limiting reactant :

0.11/2 : 0.05/1 = 0.055 : 0.05

Limiting reactant : Cl₂

mol Products based on mol Cl₂

mol products = mol HCl + mol Br₂

mol products = 2/1x0.05 + 1/1x0.05

mol products = 3 x 0.05 = 0.15

mol reactants = 0.11 + 0.05 = 0.16

<em>From Avogadro's law, </em>

In the same T, P, and V, the gas contains the same number of molecules  

So the ratio of gas volume will be equal to the ratio of gas moles  

 V₁/n₁=V₂/n₂

1 = reactants, 2 = products

Input the value :

5.85/0.16=V₂/0.15

V₂=5.484 L

6 0
3 years ago
How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC
Reptile [31]

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

4 0
3 years ago
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