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3 years ago

Write the symbol for every chemical element that has atomic number greater than 7 and atomic mass less than 24.0 u.

1 answer:

7 0

This question can best be answered using a Periodic Table of Elements. All the elements are listed by their atomic number starting from the top left and moving right and down the table. Starting at element 8, you will find Oxygen with a mass of 15.99 amu, followed by element 9 Fluorine with a mass of 18.99 , element 10 Neon with a mass of 20.18 , element 11 Sodium with a mass of 22.99. Writing them with their symbols you would have; O, F, Ne, Na.

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notka56 [123]

Answer:

417

Explanation:

50um is

50000nm

50000:120=417

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2 years ago

Please help help please

Galina-37 [17]

Answer: The correct answer is B.

Explanation: Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables.

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3 years ago

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Is the following nuclear equation balanced? Yes or No.

Maru [420]

Answer:

Reason:

During a beta decay, one neutron convert into one proton and one electron. The proton remained in nucleus while electron move away known as beta radiation. So atomic mass will be remained same but atomic no will increase by one. So correct ans will be with 218, 86 Po and beta radiation.

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3 years ago

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The air bags in automobiles were once inflated by nitrogen gas generated by the rapid decomposition of

Basile [38]

Answer:

1.57 mol NaN₃

Explanation:

2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)

First we use PV=nRT to calculate the number of N₂ moles that need to be produced:

P = 1.07 atm

V = 53.4 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K

Inputing the data:

1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86

And solving for n:

n = 2.35 mol N₂

Finally we convert N₂ moles into NaN₃ moles, using the stoichiometric coefficients of the balanced reaction:

2.35 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.57 mol NaN₃

4 0

3 years ago

A mixture of 1.374g of H2 and 70.31g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g) + Br

gregori [183]

Answer:

a. 0.139 M → [H₂] ; 0.217 M → [Br₂] ; 0.01 M → [HBr]

b. Kc = 3.31x10⁻³

Explanation:

H2(g) + Br2(g) ⇄ 2HBr(g)

Initial 1.374 g 70.31 g -

reacts X X 2x

eq. (1.374 - x) (70.31-x) 2x

In equilibrium I see, the grams I initially had minus some mass which has reacted. In products I have the double of that mass, because the stoichiometry.

So I have the mass in equilibrium, of H2 and of course I can know the mass which has reacted.

1.374g - x = 0.556 g

1.374g - 0.556 g = x = 0.808 g (This is the mass which has reacted)

70.31g - 0.808 g = 69.502 g (Mass in equilibrium of Br2)

2 . 0.808 g = 1.616 g (Mass in equibrium of HBr)

By molar mass, we can kwow the moles.

Molar mass H2: 2 g/m

Moles = mass / molar mass → 0.556 g / 2 g/m = 0.278 moles

Molar mass Br2: 159.80 g/m

Moles = mass / molar mass → 69.502 g / 159.80 g/m = 0.434 moles

Molar mass HBr: 80.9 g/m

Moles = mass / molar mass → 1.616 g / 80.9 g/m = 0.02 moles

The moles are not molarity. In equilibrium, to calculate Kc we need molarity (moles/L). The moles we have calculated are in 2 L of mixture so:

moles / L = molarity

0.278 moles / 2L = 0.139 M → [H₂]

0.434 moles / 2L = 0.217 M → [Br₂]

0.02 moles / 2L = 0.01 M → [HBr]

Kc = [HBr]² / ([H₂] . [Br₂])

Kc = 0.01² / (0.139 . 0.217) = 3.31x10⁻³

6 0

3 years ago