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Jet001 [13]
3 years ago
11

An acid-base titration is a neutralization reaction performed to determine: the concentration of an unknown acid based upon the

concentration of a known base
the concentration of an unknown acid based upon the concentration of a known acid
the amount of salt that will be produced from the reaction of the acid and base the amount of water that will be produced from the reaction of the acid and base
Chemistry
1 answer:
Inga [223]3 years ago
6 0

A is your answer, "the concentration of an unknown acid based upon the concentration of a known base"

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Answer:

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Explanation:

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3 years ago
Read 2 more answers
Please Need help asap
ruslelena [56]

Answer:

A

Explanation:

The number of protons and neutrons of an element is the same. the electrons are the only thing that can differ.  The atomic number equal the protons and neutrons.

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3 years ago
Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro
attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the <u>chemical reaction</u>:

NaN_3~->~Na~+~N_2

The next step is to <u>balance the reaction</u>:

2NaN_3~->~2Na~+~3N_2

We can continue with the <u>mol calculation</u> using the molar mass of

NaN_3 (65 g/mol), so:

3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3

Now, with the<u> molar ratio</u> between NaN_3  and N_2  we can <u>calculate the moles</u> of N_2  (2:3), so:

0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2

With the molar mass of N_2 we can <u>calculate the grams</u>:

0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2

I hope it helps!

5 0
3 years ago
A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r
slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

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3 years ago
What is understanding culture society and UCSPC?​
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Answer:

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