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4 years ago

Which of the following is a weak base ?

Chemistry

2 answers:

Scrat [10]4 years ago

7 0

<span>NH3 is the answer, if u look up its ph its a weak base</span>

inysia [295]4 years ago

5 0

Answer:

NH3 (ammonia)

Explanation:

(apex)

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Various pieces of safety equipment are used in the lab to provide protection against injury. However, the lab safety equipment which is least likely to be used when proper eye goggles are being worn is the eye wash station. If this question has the same choices as the ones posted before, the answer is letter C. eye wash station.

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4 years ago

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What is the correct name for TiF3

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Explanation:

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3 years ago

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Convert 45.5 mmHg into kPa

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3 years ago

g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.

Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0

4 years ago

The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and

WINSTONCH [101]

Answer : The molecular formula of a compound is, $C_6H_5Cl_1$

Solution : Given,

Mass of C = 64.03 g

Mass of H = 4.48 g

Mass of Cl = 31.49 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{64.03g}{12g/mole}=5.34moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.34}{0.887}=6.02\approx 6$

For H = $\frac{4.48}{0.887}=5.05\approx 5$

For Cl = $\frac{0.887}{0.887}=1$

The ratio of C : H : Cl = 6 : 5 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_6H_5Cl_1$

The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{112.5}{112.5}=1$

Molecular formula = $(C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1$

Therefore, the molecular of the compound is, $C_6H_5Cl_1$

5 0

3 years ago