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nlexa [21]
3 years ago
12

When a neutral atom gains or loses a proton, it becomes a(n)

Chemistry
1 answer:
Oksana_A [137]3 years ago
3 0

Answer:

Ion

Explanation:

I'm pretty sure yhi

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Describe the particle mode of matter. what does the brownian motion tells us about the particles in matter?give other evidence t
svetlana [45]

Answer:

dium (a liquid or a gas). This pattern of motion typically consists of random fluctuations in a particle's position inside a fluid sub-domain, followed by a relocation to another sub-domain. Each relocation is followed by more fluctuations within the new closed volume. This pattern describes a fluid at thermal equilibrium, defined by a given temperature. Within such a fluid, there exists no preferential direction of flow (as in transport phenomena). More specifically, the fluid's overall linear and angular momenta remain null over time. The kinetic energies of the molecular Brownian motions, together with those of molecular rotations and vibrations, sum up to the caloric component of a fluid's internal energy (the Equipartition theorem).

Explanation:

3 0
2 years ago
Using the redox reaction below determine which element is oxidized and which is reduced. 4nh3 + 3ca(clo)2 → 2n2 + 6h2o + 3cacl2
lys-0071 [83]
The oxidation number of elements in equation below are,

                         4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂

O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1

Oxidation:
               Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).

Reduction:
               Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).

Result:
          <span>N is oxidized and Cl is reduced.</span>
6 0
3 years ago
Read 2 more answers
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0
3 years ago
The nutritional energy content (in Calories) present in 86g of broccoli with 6g of carbohydrates, 2.6g of protein, and 0.3g of f
asambeis [7]

Answer:

37.1 calories.

Approximately, 37.1 = 40 calories.

Explanation:

So, without mincing words let's dive straight into the solution to the question above.

We are given the following parameters which are going to help in solving this particular Question.

The mass of broccoli = 86g of broccoli, mass of carbohydrates present = 6g of carbohydrates, the mass of protein present = 2.6g of protein and the mass of fat present = 0.3g of fat.

Therefore, the nutritional energy content (in Calories) = (6 × 4) + (2.6 × 4) + (0.3 × 9) = 10.4 + 24 + 2.7 = 37.1

Hence, the nutritional energy content (in Calories) = 37.1 calories.

Approximately, 37.1 = 40 calories.

5 0
2 years ago
How many grams of zinc sulfide are used to produce 1.28 grams of zinc oxide?
Lostsunrise [7]

Answer:

0.012288122055459

Explanation:

i might be wrong

6 0
2 years ago
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