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3 years ago

Which additional product balances the reaction H2SO4 + 2NaOH → Na2SO4 + ? 2H2O 2OH H2O2 H3O

Chemistry

2 answers:

Luda [366]3 years ago

7 0

Answer:

2H20 or A- the other person was right:) give them brainliest.

Explanation:

I just took the test.

Masteriza [31]3 years ago

4 0

Answer:

a) 2H2O

Explanation:

I took the quiz

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stepladder [879]

Answer:

C. element. oxygen is an element

Explanation:

oxygen is element #8

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3 years ago

This was in my chemistry class- What's the color combination code?

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2 years ago

Muscles need a stimulus from which to of the following to "tell" them contract? A. A muscle call B. The skeleton C. A nerve cell

kompoz [17]

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3 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

Answer: The volume of HCl neutralized is 1.25 mL

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of stomach acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

Hence, the volume of HCl neutralized is 1.25 mL

3 0

3 years ago

If the H+ in a solution is 1x10^-1 mol/L what is the OH-

damaskus [11]

PH scale is used to determine how acidic or basic a solution is.
we have been given the hydrogen ion concentration. Using this we can calculate pH,
pH = - log[H⁺]
pH = - log (1 x 10⁻¹ M)
pH = 1
using pH can calculate pOH
pH + pOH = 14
pOH = 14 - 1
pOH = 13
using pOH we can calculate the hydroxide ion concentration
pOH = - log [OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 10⁻¹³ M
hydroxide ion concentration is 10⁻¹³ M

8 0

3 years ago