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FromTheMoon [43]
3 years ago
8

How many elements are in the second group? (column)

Chemistry
1 answer:
zavuch27 [327]3 years ago
6 0
Of the periodic table?
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g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular
Whitepunk [10]

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called <em>neutralization reaction.</em>

The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

<em>Where the products are H2O and Ba(NO3)2</em>

4 0
3 years ago
The concentration of pb2+ in a commercially available standard solution is 1.00 mg/ml. what volume of this solution should be di
Vanyuwa [196]
<span>The concentration of pb2+ = 1.00mg/ml Diluted Solution is 6.0 x 102 ml = 612 ml Volume of the concentration of pb2+ is 0.054 mg/l is v (vL)(1.00mg/ml) = (.612L)(0.054mg/l) Volume = 0.033048L Volume of the concentration of pb2+ is 0.054 mg/l = 33.048 ml.</span>
3 0
3 years ago
There are two types of circuits we can make and they are ...
erma4kov [3.2K]
They are

parallel circuit and series circuit
4 0
3 years ago
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
A 5.5 sample of battery acid contains 490.0g of sulfuric acid H2SO4
prohojiy [21]

n =  \frac{m}{mw}  \\ n =  \frac{490}{98}  \\ n = 5 \: mol
5 0
3 years ago
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