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Sphinxa [80]
3 years ago
5

The unit used to measure barometric or air pressure

Chemistry
1 answer:
Bumek [7]3 years ago
7 0

Answer:

A barometer is an instrument used to measure air pressure and track weather systems. The most common unit of measurement used in barometers is the millibar (mb).

Explanation:

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Moles of Zn(NO3)2 in 131.50g of this substance.
Degger [83]
<span>number of moles= mass / molecular mass
mass=131.50g
molecular mass of Zn(NO3)2 =189.39 g/mol
number of moles = 131.50 g/189.39 g/mol =0.69433 mol</span>
3 0
4 years ago
Read 2 more answers
A bond in which one atom contributes both bonding electrons to a covalentbond
yKpoI14uk [10]
<h3>Answer:</h3>

A coordinate bond

<h3>Explanation:</h3>
  • A chemical bond is a force of attraction between two or more atoms of elements.
  • Covalent bond is a type of chemical bond that occurs between non-metal atoms.
  • It occurs as a result of sharing of electrons between the non-metal atoms.
  • Atoms involved in covalent bond formation may contribute equally to the bond formation or the electrons shared may come form one of the atoms.
  • When the electrons shared in a covalent bond are derived from one atom, then the covalent bond is known as a coordinate bond.
4 0
3 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
3 years ago
Please calculate the mixed volume (Vmix) when 0.300 mol of ethanol and 0.200 mol of water are mixed. Hint: The partial molar vol
Grace [21]

Answer:

If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.

Explanation:

The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.

The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:

Vmix = 0.300*57 + 0.200*17

Vmix = 17.1 + 3.4

Vmix = 20.5 cm³

7 0
3 years ago
The equilibrium constant for the reaction
FinnZ [79.3K]

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

Now put all the given values in this expression, we get :

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
3 years ago
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