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marta [7]
3 years ago
10

Who want to play among us​

Chemistry
2 answers:
lesya [120]3 years ago
8 0

Answer:

Meh i want to play

UwU

:)

zloy xaker [14]3 years ago
5 0

Answer: yeah sure lad

Explanation:

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How do you find mole ratios?
Luba_88 [7]

In order to determine the mole ratio, you need to begin with a balanced chemical equation.

3 0
3 years ago
The solubility of NaCH3CO2 in water is ~1.23 g/mL. What would be the best method for preparing a supersaturated NaCH3CO2 solutio
Len [333]

Answer:

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.

Explanation:

The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).

<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>

<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.

<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.

<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.

5 0
3 years ago
An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound
Softa [21]

Answer: Ca3(PO4)2

Explanation:

  • We assume that the sample is 100.0 g.
  • The sample contains 38.7 g of Ca, 19.9 g of P, and 41.2 g of O as the proportions show.
  • Then we can calculate the number of moles of each component (n = m/atomic mass).
  • The number of moles of Ca = 38.7/40.078 = 0.96 mol.
  • The number of moles of P = 19.9/30.97 = 0.64 mol.
  • The number of moles of O = 41.2/15.99 = 2.75 mol.
  • Now, we can get the molar ratios of different components (Ca, P, and O) in the sample by dividing the number of moles of each component by the lower number of moles, that we should divide the number of moles by (0.64).
  • Ca: P: O = (0.96/0.64) : (0.64/0.64) : (2.75/0.64) = 1.5 : 1 : 4.
  • To avoid the fraction of the ratios, we can multiply all ratios by 2.0.
  • Now, the ratio of Ca : P : O will be 3 : 2 : 8.
  • That main the empirical formula of the compound is Ca3P2O8 which can be expressed as calcium phosphate (Ca3(PO4)2).
8 0
3 years ago
How are atoms different in solids and gases?<br>​
Brrunno [24]

Answer:

In solids, the atoms are tightly packed together. In gases, atoms are spread out.

Explanation:

7 0
1 year ago
For The Reaction CH3COOH ---&gt; CH3COO- + H+, which of the following statements is true? A.) Ch3COOH is a Bronsted Lowry base.
Savatey [412]

Answer: Option (C) is the correct answer.

Explanation:

According to Arrhenius, bases are the species which when dissolved in water will give hydroxide ions, that is, OH^{-}.

For example, NaOH + H_{2}O \rightarrow Na^{+} + OH^{-}

Arrhenius acids are the species which when dissolved in water will give hydrogen ions, that is, H^{+}.

For example, CH_{3}COOH + H_{2}O \rightleftharpoons H_{3}O^{+} + CH_{3}COO^{-}

In water, when an acid loses a hydrogen ion then the specie formed is known as conjugate base.

Here, CH_{3}COO^{-} is the conjugate base of CH_{3}COOH.

Similarly, species which accept the hydrogen ion result in the formation of conjugate acid.

Hence, H_{3}O^{+} is the conjugate acid of H_{2}O.

Thus, we can conclude that CH_{3}COO^{-} is the conjugate base.

7 0
3 years ago
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