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2 years ago

Which subatomic particles affect the mass of an atom

Chemistry

1 answer:

klasskru [66]2 years ago

6 0

Answer:

protons and neutrons

Explanation:

those particles account for 99.99% of mass

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How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in

podryga [215]

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

5 0

3 years ago

A blu-ray laser has a wavelength of 405nm. In what region of the electromagnetic spectrum is this radiation? What is its frequen

love history [14]

Answer:

It lies in in the visible region.

Frequency: is 7.407 × 10^-14 Hz

${ \bf{V = f \lambda}} \\ { \tt{3.0 \times {10}^{8} = f \times (405 \times {10}^{ - 9}) }} \\ { \tt{f = 7.407 \times {10}^{14} \: hertz}}$

Explanation:

V is speed of light.

f is frequency

lambda is the wavelength

8 0

2 years ago

SOMEONE PLEASE HELP ME! MEDAL, FAN, AND TESTIMONIAL

yKpoI14uk [10]

As we can see the chemical equation is balanced.K3PO4 + Al(NO3)3 → 3KNO3 + AlPO4

So, by principle of conservation of mass when 1 mole of K3PO4 reacts with 1 mol of Al(NO3)3 , it peoduces 3 mol of KNO3 and 1 mol of AlPO4

So, when 2.5 moles of potassium phosphate react and Al(NO3)3 is present in excess , 2.5*3= 7.5 mol of KNO3 is formed

8 0

3 years ago

6) What is the electron configuration of an atom in its ground state with a

shutvik [7]

Answer: the answer is B!

Explanation:

S and p’s are valence electrons and if added, there is 7 in total !

8 0

2 years ago

Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).

babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

8 0

2 years ago