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dusya [7]
3 years ago
5

Which substance is the limiting reactant when 4.0 g of sulfur reacts with 6.0 g of oxygen and 8.0 g of sodium hydroxide accordin

g to the following chemical equation: 2 S(s) + 3 O2(g) + 4 NaOH(aq) → 2 Na2SO4(aq) + 2 H2O(l)
A. O2(g)
B. S(s)
C. NaOH(aq)
D. None of these substances is the limiting reactant.
Chemistry
1 answer:
Marysya12 [62]3 years ago
6 0
C, NaOH(aq) is the answer
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Calculate the pHpH of a 0.10 MM solution of HClHCl . Express your answer numerically using two decimal places.
alisha [4.7K]

Answer: The pH 0f 0.10 solution of HCl is 1.00

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,  

1 mole of HCl gives 1 mole of H^+  

Thus 0.10 moles of HCl gives =\frac{1}{1}\times 0.10=0.10 moles of H^+  

Putting in the values:

pH=-\log[0.10]

pH=1.00

Thus pH 0f 0.10 solution of HCl is 1.00

8 0
3 years ago
How are the African plate and south American plate moving relative to each other
Kamila [148]
They are moving away from each other
6 0
3 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
At what temperature does uranium hexafluoride have a density of 0.9560 g/L at 0.5073 atm?
S_A_V [24]
Given:



Density = 0.7360 g/L.

Pressure = 0.5073 atm.
Step 2
The mathematical expression of an ideal gas is,

Chemistry homework question answer, step 2, image 1
Step 3
Here, R is the universal gas constant (0.0821 L-atm/mol K), T is the temperature in Kelvin, and n is the number of
8 0
3 years ago
(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM
torisob [31]

Answer:

The answer is "20 \ mL"

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\  Tris\\\\

therefore,

it takes 20 mL of Tris.

\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\

                     = 30 \ mL \ of\ Nacl

So, take 20 \ mL\ of\ NaCl.

6 0
3 years ago
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