Which ingredients produce the best bar for growth and repair?

How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in

podryga [215]

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

5 0

3 years ago

What is the pH of a solution that has a hydrogen ion concentration of 1.0 * 10 -9 M? Show your calculation.

Nastasia [14]

Answer:

9.00

Explanation:

Data:

[H⁺] = 1.0 × 10⁻⁹ mol·L⁻¹

Calculation:

pH = -log[H⁺] = -log(1.0 × 10⁻⁹) = -log(1.0) - log(10⁻⁹) = -0.00 - (-9) = -0.00 + 9 = 9.00

7 0

2 years ago

Separating mixtures: filtration

Ymorist [56]

hi these are the answers :)

5 0

1 year ago

Liquid nitrogen is cold and can be used to cool objects to -196°C. If you put the bottle of warm air in liquid nitrogen at this

vampirchik [111]

Answer:

Explanation:

It depends on how this is done. If you raise the pressure, the nitrogen will disappear (liquify) and all that will be left will be the 21 % oxygen and the 1% argon.

The process is very complicated because the boiling point of nitrogen keeps on changing. The boiling point is unstable.

6 0

2 years ago

To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water

Masteriza [31]

Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0

3 years ago