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Liono4ka [1.6K]
3 years ago
8

Hi, can someone help me balance this chemistry equation: H2SO4 + RbOH -> Rb2SO4 + H2O

Chemistry
1 answer:
zloy xaker [14]3 years ago
6 0

H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

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x = (13g*1mol) / 0,26mol =  50g

It could be Ti or V
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What is the product of the reaction between arsenic and oxygen
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d

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Which is a common way for a scientific calculator to show the number below? <br> 8.1 102
Elodia [21]
Answer: 8.1e+2 or 8.1E+2 (in the scientific notation mode), based on that the number given is 8.1 × 10².

Explanation


Scientific calculators use the letter e or E to show the numbers in scientific notation mode.

These are some examples of numbers that require scientific notation and how they are shown.

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6.022 × 10²³           6.022 e+23
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So, understanding that your question is about number 8.1 × 10², the scientific calculator could show  8e+2.

Nevertheless, since 8 .1 × 10² is 810, not a long number, if the display is not in scientific notation mode, it could just display just 810,

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3 years ago
A substance is matter that has a definite composition.<br> TRUE or FALSE?
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An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
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Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

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Best regards!

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