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JulsSmile [24]
3 years ago
10

Bacteria can be heterotrophs or autotrophs true or false?

Chemistry
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

True

Explanation:

I learned about it a few years ago

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What type of bond is formed between hydrogen and nitrogen in ammonia (NH3)? A. polar covalent B. ionic C. nonpolar covalent D. m
cricket20 [7]
The answer is A polar covalent
5 0
3 years ago
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11. A new compound has just been developed. Its formula is:
jarptica [38.1K]

Answer:

6.

Explanation:

G, H, O, N, Na, P

There are 6 different elements listed, with O (oxygen) showing up twice.

7 0
3 years ago
How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o
Sladkaya [172]

Potassium 23.5g/39.0983g/mol = 0.601mol

The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol

Therefore 0.3005mol of F2 is needed to find liters use

formula V = nRT/P (V)Volume = 22.41L

(T)Temperature = 273K or 0.0 Celsius

(P)Pressure = 1.0atm

<span>(R)value is always .08206 with atm n = 0.3005moles (273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>

8 0
3 years ago
Read 2 more answers
Help, please? will mark brainliest​
alexira [117]

Answer:

Sulfur would gain electrons

Explanation:

Atoms want to have a complete out valence shell and because sulfur only needs 2 more electrons to complete the outer shell it would take 2 more.

8 0
3 years ago
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04
evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0
3 years ago
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