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guapka [62]
3 years ago
7

How many moles of tin, Sn, are in 2500 atoms of tin?

Chemistry
1 answer:
mario62 [17]3 years ago
6 0
<h3>Answer:</h3>

4.2 × 10⁻²¹ moles Sn

<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>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

2500 atoms Sn

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2500 \ atoms \ Sn (\frac{1 \ mol \ Sn}{6.022 \cdot 10^{23} \ atoms \ Sn})
  2. Multiply:                            \displaystyle 4.15144 \cdot 10^{-21} \ moles \ Sn

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

4.15144 × 10⁻²¹ moles Sn ≈ 4.2 × 10⁻²¹ moles Sn

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A helium filled weather balloon has a volume of 806 L at 20.9°C and 753 mmHg. it is released and rises to an altitude of 6.8 km,
OLEGan [10]
<h3>Answer:</h3>

1257.45 L

<h3>Explanation:</h3>

We are given;

  • Initial volume of Helium gas, V1 as 806 L
  • Initial temperature of Helium gas,T1 as 20.9°C
  • Initial pressure of Helium gas, P1 as 753 mmHg
  • Pressure of Helium at the altitude 6.8 km, P2 as 417 mmHg
  • Temperature of Helium gas at the altitude 6.8 Km, T2 as -19.1°C

But, K = °C + 273.15

Therefore, T1 = 294.05 K and T2 = 254.05 K

  • We are required to calculate the new volume of the balloon at 6.8 km.
  • To determine the new volume we are going to use the combined gas law.
  • According to the combined gas law, \frac{P1V1}{T1}=\frac{P2V2}{T2}

Thus, rearranging the formula;

V2=\frac{P1V1T2}{P2T1}

V2=\frac{(753)(806L)(254.05K)}{(417)(294.05)}

V2=1257.45L

Therefore, the volume of the balloon at an altitude of 6.8 km is 1257.45 L

7 0
3 years ago
Who discovered electrons using Crookes tube?
andrey2020 [161]

Answer:

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A 135 g sample of a metal requires 2.50 kJ to change its temperature from 19.5°C to 95.5°C to 100.0°C?
just olya [345]

Answer:

0.23J/g°C

Explanation:

Given parameters:

Mass of sample  = 135g

Amount of heat  = 2.5kJ

Initial temperature  = 19.5°c

Final temperature  = 100°C

Unknown:

Specific heat capacity of the metal  = ?

Solution:

The specific heat capacity of a substance is the amount of heat required to the raise the temperature of 1g of the substance by 1°C.

   H  = m C (T₂ - T₁ )

H is the amount of heat

m is the mass

C is the specific heat capacity

T₂ is the final temperature

T₁ is the initial temperature;

     2.5 x 10³  = 135 x C x (100 - 19.5)

     2500 = 10867.5C

       C = \frac{2500}{10867.5}   = 0.23J/g°C

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Convection currents are caused by the movement of the magma in the mantle.

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