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Simora [160]
3 years ago
13

When looking at flat-tailed horned lizards, what trait was found in most living lizards compared to dead lizards? *

Chemistry
2 answers:
UkoKoshka [18]3 years ago
5 0

Answer:

d after any reptile that is poisinis when it dies the amount of poisin becoms weaker becuase its rrying to kill air but it kill nothing

Explanation:

tensa zangetsu [6.8K]3 years ago
4 0
B the horn was longer
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Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your
Minchanka [31]
When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

      = 85 J
5 0
3 years ago
How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *
yuradex [85]
<h3>Answer:</h3>

\displaystyle 0.1 \ mol \ Fe

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.022 × 10²² atoms Fe (iron)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})
  2. Divide:                    \displaystyle 0.1 \ mol \ Fe
7 0
3 years ago
5. Why doesn't the air at a frontal boundary mix?​
Vsevolod [243]

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

4 0
3 years ago
There are some exceptions to the trends of first and successive ionization energies. For each of the following pairs, explain wh
NNADVOKAT [17]

Answer:

(b) IE₂ of Ga >  IE₂ of Ge

Explanation:

Electronic configuration of Ga is [Ar] 3d¹⁰4s²4p¹

Electronic configuration of Ge is [Ar] 3d¹⁰4s²4p²

After 1st ionisation , Ga  becomes [Ar] 3d¹⁰4s² and becomes stable . Its

2 nd ionisation requires higher amount of ionisation energy. In case of Ge , there are 2 electrons in outermost orbital so it becomes stable after ionisation of 2 electrons.

4 0
3 years ago
Read 2 more answers
Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
drek231 [11]

Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

6 0
3 years ago
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