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katovenus [111]
2 years ago
11

What does it mean if you titrated the acid to a 'hot pink' color All of these

Chemistry
1 answer:
Murrr4er [49]2 years ago
3 0

Answer: Too much base was added

i guessed

Explanation:

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Can someone help me write chemical equations of what went wrong in the Chernobyl nucleur accident? Eg. graphite in the rods spee
zimovet [89]

Answer:

It might be e=FCA

Explanation:

4 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
Bromine has a density of 3.10g/cm3. If you have 50.0 ML of bromine, how many grams do you have?
Zanzabum

Answer:

The answer is

<h2>155 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of bromine = 50 mL

density = 3.10 g/cm³

It's mass is

mass = 50 × 3.10

We have the final answer as

<h3>155 g</h3>

Hope this<u> </u>helps you

5 0
3 years ago
Read 2 more answers
Please help I will give brainiest
Sergio [31]

Answer:

Answer

Answer is option B coz da negative charge r more thn positive

8 0
3 years ago
Read 2 more answers
1. Energy changes occur as *
alexandr1967 [171]
Number 3 i think is <span>d.heat moves from an object of higher temperature to an object of lower temperature</span>
3 0
3 years ago
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