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ololo11 [35]
3 years ago
14

How many kilograms is 9 grams

Chemistry
2 answers:
Lubov Fominskaja [6]3 years ago
7 0

9000 kilograms are in 9 grams

eduard3 years ago
3 0

The answer is 0.009

hope this helps

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Identify the item that does not have kinetic energy.
d1i1m1o1n [39]

Answer:

D

Explanation:

there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting

4 0
3 years ago
An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the am
gogolik [260]

Answer:

Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

Explanation:

4 0
3 years ago
You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions
artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

8 0
3 years ago
Hellllppppp<br> lons Worksheet
Iteru [2.4K]

The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.

An ion can be cation (positively charged) or anion (negatively charged).

Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).

The remaining elements are completed as follows to attain octet structure;

<u>Element</u>--<u>valence electron</u>--<u>electrons to gain</u>--<u>electrons to lose</u>--<u>ion formed</u>

O ------------ 6 ----------------------  2 ------------------------  none -------------- O^{2-}

Ca --------   2 ----------------------- none ---------------------- 2 ------------------ Ca^{2+}

Br ----------- 7 ---------------------     1 ------------------------ none --------------- Br^{-}

S ------------ 6 -----------------------  2 ------------------------ none --------------- S^{2-}

Cl ------------ 7 -----------------------  1 ------------------------ none ----------------Cl^{-}

K -------------- 1 ----------------------- none ----------------------- 1 ------------------ K^{+}

Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- Mg^{2+}

Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- Be^{2+}

Learn more here: brainly.com/question/21089350

8 0
3 years ago
At 700 K, the reaction 2SO2(g) + O2(g) &lt;====&gt; 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f
nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

3 0
3 years ago
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