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3 years ago

Which compounds have the empirical formula CH2O?

Chemistry

2 answers:

lara [203]3 years ago

8 0

Answer:

The answer to your question is all the formulas in bold has the same empirical formula

Explanation:

Data

Empirical formula CH₂O

Process

To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.

a) C₂H₄O₂ factor 2 2(CH₂O)

b) C₃H₆O₃ factor 3 3(CH₂O)

c) CH₂O₂ this formula can not be simplified

d) C₅H₁₀O₅ factor 5 5(CH₂O)

e) C₆H₁₂O₆ factor 6 6(CH₂O)

GenaCL600 [577]3 years ago

6 0

Answer:

C2H4O2

C3H6O3

C5H10O5

C6H12O6

Explanation:

CH2O2 can not be broken down or simplified

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A piston contains 0.39L of air at 1.0 atm. If the pressure on the piston is increased to 5.6 atm, what is the new volume?

bazaltina [42]

The new volume : 0.07 L

<h3>Further explanation</h3>

Boyle's Law

At a constant temperature, the gas volume is inversely proportional to the pressure applied

$\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

P₁=1 atm

V₁=0.39 L

P₂=5.6 atm

$\tt V_2=\dfrac{P_1.V_1}{P_2}\\\\V_2=\dfrac{1\times 0.39}{5.6}\\\\v_2=\boxed{\bold{0.07~L}}$

3 0

3 years ago

Please and thank you

Virty [35]

3.8mL of 0.42 phosphoric acid is required.

Reaction

2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl

moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol

moles of H3Po4

= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2

= 0.00106 mol

V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL

V of H3PO4=3.8mL

To know more about calculation in milliliters refer to:-

brainly.com/question/23276655

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4 0

2 years ago

Elements in the same period of the periodic table exhibit similar physical and chemical properties.

Vladimir [108]

Answer:

same valency electrons

Explanation:

example g 1 elements

5 0

3 years ago

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Westkost [7]

Answer:

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Explanation:

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4 0

3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0

3 years ago