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Law Incorporation [45]
3 years ago
11

How would you use density to identify an unknown substance

Chemistry
2 answers:
alexdok [17]3 years ago
8 0
Seaport is correct for your question


ale4655 [162]3 years ago
4 0
You would measure the substances mass and its volume, then divide the mass by the volume.
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Please help me with these Radioactive Decay Problems ASAP! I need helppppp
Masteriza [31]

Answer:

5

Explanation:

it is very simple

5 0
3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘c is 2000 m/s. Note that 1. 0 mol of diatomic hydrogen at
Leno4ka [110]

The rms speed will be 500 m/s

<h3>What is Root mean square speed ?</h3>

Root mean square speed (Vrms) is defined as the square root of the mean of the square of speeds of all molecules.

Root mean square speed (vrms) Root mean square speed (vrms) is defined as the square root of the mean of the square of speeds of all molecules

It is given that

Speed of a diatomic hydrogen molecule,2000 m/s

Mol of diatomic hydrogen,1.0

Temperature,50°C

The rms speed of diatomic molecule will be:

√(3KT)/( m)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

(3/2)KT = (1/2) mv²

v = √(3KT)/m  

FOR H₂:  v = √(3KT)/m = 2000 m/s  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

Therefore the  rms (root-mean-square) speed of a oxygen molecule at 50∘c is 500m/s.

To know more about Root mean square speed

brainly.com/question/7213287

#SPJ4

3 0
2 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
Draw structure of oxygen, nitrogen &amp;methane.​
Elina [12.6K]

See the image below-⬇⬇-

Hope this helps :D

4 0
3 years ago
Read 2 more answers
45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution
Stells [14]
As we know that Molarity is given as,

                                       M = moles / V 
Solving for V,
                                        V = moles / M ------------------(1)
Also, moles is equal to,
                                       moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
                                       V = (mass / M.mass) / M
Putting values,
                                       V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³ 
6 0
3 years ago
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