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Mumz [18]
3 years ago
9

An element has an atomic number of 29. The number of protons and electrons in a neutral atom of the element are *

Chemistry
1 answer:
nasty-shy [4]3 years ago
8 0

<em>The number of protons and electrons in a neutral atom of the element are;</em>

D. 29 protons and 29 electrons

<u>The atomic number of an element is equal to the number of protons in the nucleus of each atom of that element.</u>

<u>Thus copper has an atomic number of 29, all atoms of copper will have 29 protons.</u>

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using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius​
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Answer:

Please, see attached two figures:

  • The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

  • The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

  • Answer: <u>5.5g</u>

Explanation:

The red  arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

            10mL\times 1.0g/mL=10g

Thus, the solutibily is:

      10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl

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Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation: at 250° 0.125 m pc
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7) For the reaction 9A (g) + B (g)  5C(g) + 1/6 D (g), it takes 4 and a half minutes for the concentration of C to increase to
viva [34]

Answer: The correct option is, (C) 0.53

Explanation:

The given chemical reaction is:

9A(g)+B(g)\rightarrow 5C(g)+\frac{1}{6}D(g)

The rate of the reaction for disappearance of A and formation of C is given as:

\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}

Or,

\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}

where,

\Delta C = change in concentration of C = 1.33 M

\Delta t = change in time = 4.5 min

Putting values in above equation, we get:

\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}

\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}

\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}

\frac{\Delta [A]}{\Delta t}=0.53M/min

Thus, the decrease in A during this time interval is, 0.53

5 0
3 years ago
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