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ololo11 [35]
3 years ago
15

Help on #6 please!! Chemistry

Chemistry
1 answer:
murzikaleks [220]3 years ago
3 0
It's just asking for 5% or 31 grams
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The molecule CH,Cl has four atoms arranged around one central atom. What is the shape of the molecule? (1
balu736 [363]

Answer:

tetrahedral

Explanation:

it sounds like a tetrahedral shape, which is when there are four bonds and no lone pairs around the central atom in the molecule.

6 0
2 years ago
Does hydrogen builds all the acids??​
Umnica [9.8K]

Answer:

hydrogen builds many acids but not all

6 0
3 years ago
The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for
Mashcka [7]

Answer:

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants  2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

3 0
3 years ago
How many total electrons does the P^3- ion have? O a. 3 O b. 31 O c.1 O d. 15 O e. 18
STALIN [3.7K]

Answer:

e. 18

Explanation:

A neutral P atom has an atomic number of 15, which means there are 15 protons in the atom. In order to be neutral, the P atom must also have 15 electrons.

The P³⁻ anion has 3 electrons more than the neutral P atom since it has a charge of -3.

Thus, the total number of electrons are 15 + 3 = 18 electrons.

7 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
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