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kiruha [24]
2 years ago
15

in this trial, 0.400 M NaOH was added to 40.00 ml of 0.400 M HCl. How many ml of base must be added to cause the colour to chang

e?​
Chemistry
1 answer:
denis-greek [22]2 years ago
6 0

Answer:

The amount of Ml that is needed

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Where in the laboratory is long, unrestrained hair most likely to be a safety concern? near an open flame close to a fume hood n
zimovet [89]

Answer:

near an open flame

Explanation:

if your hair is not tied back near open flames in combination with most hair care products it would turn you into a living torch

7 0
2 years ago
How does the electron structure of atoms change when they form chemical bonds
andriy [413]
For ionic bond
The metal atom will lose electrons to form cations and the non metal atom will gain electron.egNaCl
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2 years ago
What is the volume of silver metal will weigh exactly 2500.0g. The density of silver is 10.5g/cm3 -centimeters cubed-
Lubov Fominskaja [6]

Answer:

volume = 238.09 cm³

Explanation:

see the file

7 0
3 years ago
Could a mixture be made up of only one element and no compounds​
Firdavs [7]

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5 0
2 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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