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2 years ago

15

A pH of 2 indicates a(n) _____.

1 answer:

solniwko [45]2 years ago

6 0

A pH of 2 indicates a acid

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What step is ussaly done right before the experimentation in the scientific method

barxatty [35]

The Step that is Usually done right before the experimentation in the Scientific Method is Developing a Hypothesis and asking a Question.

Hope that helped!

~Izzy

3 0

2 years ago

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Decomposition of hydrogen peroxide: 2H2O2 --> O2(g) + 2H2O(l) How many molecules of water are produced from the decomposition

Lubov Fominskaja [6]

Answer:

Explanation:

You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass

<em>3.4 g/2.016 = 1.686 moles.</em>

Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.

1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.

Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So

1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.

8 0

1 year ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

2 years ago

Give two reasons why quarrying is bad for the enviroment

Elza [17]

Air & noise pollution
Quarry waste
Damage to biodiversity

6 0

2 years ago

C2h6o how many moles of ethanol are present in a 10.0 g sample of ethanol

Dmitriy789 [7]

46 gram of ethanol ≡ 1 mole of ethanol
1 gram of ethanol ≡ 1/46 mole of ethanol
10 gram of ethanol ≡ 1*10/46 mole of ethanol
=0.217 mole of ethanol

5 0

2 years ago

Read 2 more answers