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maxonik [38]
3 years ago
11

Which has a greater amount of particles, 1.00 mole of hydrogen (H) or 1.00 moles of oxygen(0)?

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

Answer:

Both have the same amount of particles.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.

This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.

Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.

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Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?
Vedmedyk [2.9K]
The ideal gas law may be written as
p= \frac{\rho R T}{M}
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
   = 221.95 kPa 
   = (2.295 x 10⁵)/101325 atm
   = 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

4 0
3 years ago
Someone help me with question pls
gizmo_the_mogwai [7]

Explanation:

the conductors are the three u have checked

8 0
3 years ago
Which of the following statements is (are) true about enzyme-catalyzed reactions? a. The reaction is faster than the same reacti
Brums [2.3K]

Answer : The correct option is A.

Explanation :

Enzyme-catalyzed reaction :

Enzyme act as a biological catalyst and the role of catalyst is to increase the rate of chemical reaction by lowering the activation energy.

Most of the chemical reactions are slow in the absence of enzyme but in the presence of enzyme, the reaction become faster. That means the Enzyme accelerate the rate of reaction.

Therefore, the correct answer is the reaction is faster than the same reaction in the absence of the enzyme.

5 0
3 years ago
The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti
lyudmila [28]

NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}

Here the base is a benzoate ion, which is a weak base and reacts with water.

C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or [OH^{-}] = 10^{^{-pOH}}

[OH^{-}] = 10^{^{-4.96}}

[OH^{-}] = 1.1\times 10^{-5}

The base dissociation equation kb = \frac{Product}{Reactant}

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = 1.6\times 10^{-10}

And value of [OH-] we have calculated as 1.1\times 10^{-5} and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}

[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}

[C6H5COO^{-}] = 0.76 M or 0.76\frac{mol}{L}

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = 0.76(\frac{mol}{L}) \times (0.50L)

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0
3 years ago
Read 2 more answers
A 25 Ω resistor has a voltage drop of 12 V across it. Calculate the current flowing through the resistor
Yanka [14]

Answer:

0.48

Explanation:

i =  \frac{v}{r}

because we are looking for I which is current we say

I = 12÷25 which is 0.48.

4 0
3 years ago
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