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Dimas [21]
3 years ago
13

Name the SI base units that are important in chemistry, and give the SI units for expressing the following: (a)length, (b) volum

e, (c) mass, (d) time, (e) temperature.
Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer and Explanation:

The basic unit which are that are important in chemistry are meter, kilogram ,mol,m^3

Candela which is the unit of luminous of intensity is not so important in physics

(a) SI unit of length is meter (m)

(b) Si unit of volume is m^3

(c) Si unit of mass is kilogram (kg)

(d) SI unit of time is second (s)

(e) SI unit of temperature is kelvin (K)

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What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO
docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

  • \rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0
3 years ago
How many moles are there in 2.30 x 1024 atoms of silver?
Sunny_sXe [5.5K]

Answer:

Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).

Step 1

Use dimensional analysis to convert the number of atoms to moles.

1 mole atoms = 6.022 × 10²³ atoms

n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag

Step 2

Convert the moles of Ag to mass.

mass (m) = moles (n) × molar mass (M)

n(Ag) = 3.8193 mol Ag

M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag

m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures

The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.

Explanation:

5 0
3 years ago
What is the net force of 4N 2N!!
RSB [31]

Answer: 4 N-2 N=2 Nto the left. Change in motion: The box will move to the left. 7.)

Explanation: Please mark branliest.

8 0
3 years ago
Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and
IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x.  So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0
3 years ago
Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0
4 years ago
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