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xeze [42]
3 years ago
12

I need help with this please

Chemistry
1 answer:
Annette [7]3 years ago
8 0

Answer:

1=Rb, 2=O, 3=As, 4=Al and 5=Ca

Explanation:

You might be interested in
How many protons and neutrons are there in an atom of 3.K?
pickupchik [31]

Answer:

A: 11 protons and 21 neutrons

Explanation:

Hopefully this helps!

3 0
3 years ago
A frying pan needs a Teflon coating of 1.00 mm in thickness to cover an area of 36.0 square inches. How many ounces of Teflon ar
Otrada [13]

Answer:

m = 0.659 ounce

Explanation:

It is given that,

The thickness of a Teflon coating is, d = 1 mm

Area of the coating, A = 36 inch²

The density of Teflon, d = 0.805 g/mL

We need to find ounces of Teflon are needed.

Firstly, find the volume of the Teflon needed,

1 inch² = 6.4516 cm²

36 inch² = 232.258 cm²

Density,

\rho=\dfrac{m}{V}

V is volume of the Teflon needed, V = Ad

So,

m=\rho V\\\\m=0.805\ g/cm^3\times 232.258\ cm^2 \times 0.1\ cm\\\\m=18.69\ g

Also, 1 gram = 0.035274 ounce

18.69 gram = 0.659 ounce

So, 0.659 ounces of Teflon are needed.

3 0
3 years ago
Read temprature on thermometer i am confusion
balu736 [363]

Answer:

a)22.2°C after adding magnesium

b)17.3°C before adding magnesium

c) 4.9 is change

8 0
3 years ago
What type of ions do nonmetals naturally form?
tensa zangetsu [6.8K]
Non-metal atoms gain an electron, or electrons, from another atom to become >negatively charged ions.
8 0
3 years ago
Read 2 more answers
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
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