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umka2103 [35]
3 years ago
10

To support a child's self-esteem, you can make all of these statements, except

Chemistry
1 answer:
mihalych1998 [28]3 years ago
7 0
The answer is A you shouldn't discourage a child like that

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A compound has an empirical formula of CH2 and a molar mass of 56 g. What is its molecular formula?
kipiarov [429]

Answer:

C4H8

Explanation:

First find the molar mass of CH2;

2(1.01) + 1(12.01) = 14.03g

Now divide the molar mass of the compound by the molar mass of CH2;

56g/14.03g = 3.9914 Round to nearest whole number = 4

Multiply CH2 by 4 to get the molecular formula;

CH2* 4 = C4H8

4 0
3 years ago
The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M , [ H + ] = 2.00 × 10 − 4 M [H+]=
ValentinkaMS [17]

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

    = 15.38 x 10⁻⁸

Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

7 0
3 years ago
if you find that a substance turns litmus paper to blue and contains hydroxide ions what would it most lightly be
Delicious77 [7]
Acid if I remeber correctly
5 0
3 years ago
Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

3 0
3 years ago
Are all physical reactions reversible?
LUCKY_DIMON [66]

Answer: answer is A

Explanation:

6 0
3 years ago
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