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AnnZ [28]
3 years ago
12

Positive feedback is when the change caused by the stimulus is A )decreased B) changes constantly C) nonexistent D) Increased HE

LP A.S.A.P
Chemistry
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

c) nonexistent

hope it helps

plz mark as brainliest

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Sodium azide, NaN3, the explosive compound found in automobile air bags, decomposes according to the following equation: 2NaN3(s
shutvik [7]

Answer:

1.9 × 10² g NaN₃

1.5 g/L

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles of N₂ formed

N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol

We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.

4.2 mol × 28.01 g/mol = 1.2 × 10² g

Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂

The mass ratio of NaN₃ to N₂ is 130.02:84.03.

1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃

Step 4: Calculate the density of N₂

We will use the following expression.

ρ = P × M / R × T

ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L

5 0
3 years ago
Please help with this question ;/
aev [14]
The link above is a hacker
8 0
3 years ago
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AnnZ [28]
Here you are looking on the Free Body diagram of a net force of 0N in both the x and y-directions.  the only ones that has that condition met is A and C.
4 0
3 years ago
What is the ionic charge for the calcium ion in CaCO3?
Lunna [17]
<span>The ionic charge of Calcium (Ca) in calcium carbonate (CaCO3) is 2+. CaCO3 has a neutral ionic charge sin CO3 has a 2- charge.</span>
3 0
3 years ago
The temperature of 15.71 grams of gold rises from 32°C to 1,064°C, and then the gold melts completely. If gold’s specific heat i
antoniya [11.8K]

Answer:Total energy gained by 15.71 g is 3090.6471 joules

Explanation:

Given:

Q = heat gained by the 15.71 gram mass of gold

Q=mc\Delta T +m\Delta H_{fusion}

\Delta T=(T_{final}-T_{initial})=(1064^oC-32^oC)=1032^oC

c = specific heat capacity of gold = 0.1291joules/gram^oC

m = mass of gold =15.71 g

Q=15.71 g\times 0.1291 joules/gram^oC (1032^oC) +15.71 g\times 63.5 Joules/gram

Heat gained by gold = 2093.0621 Joules + 997.585 Joules

=3090.6471 joules

6 0
3 years ago
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