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3 years ago

Infer: Is the reaction below possible? Explain why or why not. H2O + NaOH NaCl + H2

1 answer:

3 0

Answer: No because of the equation Hydrogen is brought out of it. This will make the equation uneven therefor it is invalid.

Explanation:

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1. The maximum recommended amount of citric acid (H3C6H5O7; the 3 hydrogens listed first are the acidic hydrogens – the only hyd

Lisa [10]

The density has been given as the mass of the substance per volume of the solution. The mass of citric acid in the bottle has been 1.22 grams.

<h3>What is percent mass?</h3>

Percent mass can be given as the mass of substance in the total mass of the solution.

The volume of the face serum has been 250 mL, and the density is 0.979 g/mL.

The mass of face serum in the bottle has been:

$\rm Density=\dfrac{Mass}{Volume}\\ 0.979\;g/mL=\dfrac{Mass}{250\;mL}\\ Mass=244.75\;g$

The total mass of face serum has been 244.75 grams.

The percent mass of citric acid in face serum is 0.5%. This states the presence of 0.5-gram citric acids in 100 grams sample.

The mass of citric acid in the face serum sample is given as:

$\rm 100\;gram\;sample=0.5\;g\;H_3C_6H_5O_7\\\\244.75\;grams\;sample=\dfrac{0.5}{100}\;\times\;244.75\;grams\; H_3C_6H_5O_7\\\\244.75\;grams\;sample=1.22\;grams\;H_3C_6H_5O_7$

The mass of citric acid present in 250 mL serum sample has been 1.22 grams.

Learn more about percent mass, here:

brainly.com/question/5394922

3 0

2 years ago

Which of these is an example of water in liquid form

skad [1K]

Answer:

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4 0

3 years ago

Read 2 more answers

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

Methylpropane (complete and condensed structural formulas)

Tcecarenko [31]

Answer:

Here's what I get

Explanation:

1. Complete structural formula

Methylpropane consists of a chain of three carbons with another carbon atom attached to the middle carbon. Enough H atoms are added to give each C atom a total of four bonds.

The complete structural formula is shown below (There is a C atom at each intersection).

2. Condensed structural formula

A condensed structural formula is designed to be typed on one line.

The molecule has three CH₃ groups attached to a single carbon atom, so the condensed structural formula is

(CH₃)₃CH

The formula is also often written CH₃CH(CH₃)CH₃ and as (CH₃)₂CHCH₃.

5 0

3 years ago

→

marin [14]

Answer:

[A]²

Explanation:

Since the formation is independent of D, D is 0 order.

Since a quadruples when it is doubled it can be written as

2A^X= 4

To find the unknown power we can assume A= 1 to make the math simple. So When a = 2 (Because you doubled it) raised to X power it will equal 4

so the unknown power is 2

Making the rate law

[a]²[b]⁰

or simply just

[A]²

8 0

3 years ago