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Alona [7]
3 years ago
6

An element has the following subatomic particles: 8 protons, 8 neutrons, 8 electrons. What element is it?

Chemistry
2 answers:
Marina86 [1]3 years ago
8 0
Oxygen is the answer. it has 8 neutrons and 8 electrons as well as 8 protons.
Yuliya22 [10]3 years ago
3 0
The answer to your question is oxygen
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What is the correct way to write 1,550,000,000 in scientific notation
aliina [53]
It would be 1.55x10 to the 9th
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6 0
3 years ago
The formula for chlorine acid is
fenix001 [56]

The correct terminology for chlorine acid is Hydrochloric Acid. The formula for that is HCl. This is because H has an ionic charge of +1 and Cl has an ionic charge of -1. Thus, they will combine to form HCl. Also, H is in front because almost all acids will have H in front.

5 0
3 years ago
To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r
svetoff [14.1K]

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂   →  CO₂

Balanced chemical equation:

2CO + O₂   →  2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂  = 0  kj/mol

The standard enthalpy for the formation of CO₂  = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

7 0
3 years ago
A balloon has a volume of 125mL with 10 pumps of gas. The balloon is reduced in volume to 88mL, how much gas is in the balloon?
blagie [28]

Answer:

7.04

Explanation:

i think

6 0
3 years ago
At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the
Aleks [24]

Answer:

K_{p}=4.35\times 10^{-4}

Explanation:

We know, K_{p}=K_{c}(RT)^{\Delta n}

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and \Delta n is difference in sum of stoichiometric coefficient of products and reactants

Here \Delta n=(2)-(2+1)=-1 and T = 311 K

So, K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}

Hence value of equilibrium constant in terms of partial pressure (K_{p}) is 4.35\times 10^{-4}

4 0
3 years ago
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