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const2013 [10]
3 years ago
12

What number minus 6 equals 58?

Chemistry
2 answers:
Novay_Z [31]3 years ago
3 0

Answer: 64

Explanation:

goldenfox [79]3 years ago
3 0

Answer:

64

Explanation: add 6 to 58 and you get 64.

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Calculate the half-life of different radioactive sources.
otez555 [7]
The rate at which a radioactive<span> isotope decays is measured in </span>half-life. The termhalf-life<span> is defined as the time it takes for one-</span>half<span> of the atoms of a </span>radioactive material<span> to disintegrate. </span>Half-lives<span> for </span>various radioisotopes<span> can range from a few microseconds to billions of years.
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zane
7 0
3 years ago
How are oxidation-reduction reactions related to how you use energy?​
Alexxandr [17]

Answer:

In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. The movement of electrons can be used to do work. ... The electron flow can be run through a wire and these electrons can be used to do work (like run a battery). Hope this helps.

3 0
3 years ago
Urgent‼️‼️‼️‼️‼️‼️ Find the average atomic mass of element Uus. 284Uus 283.4 amu 34.60 % 284.7 amu 21.20 % Uus 287.8 amu 44.20 %
Citrus2011 [14]

Answer:

285.6

Explanation:

(283.4×0.3460)+(284.7×0.2120)+(287.8×0.4420)

=285.6204

4 0
3 years ago
If 28 ml of 5.4 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?
Ainat [17]

A reaction between acid and base to form water and salt is known as neutralization reaction. It is a double replacement reaction.  

The reaction between H_2SO_4 and NaHCO_3 will be:

H_2SO_4 + NaHCO_3 \rightarrow Na_2SO_4 + CO_2 + H_2O

The balanced reaction is:

H_2SO_4 + 2NaHCO_3 \rightarrow Na_2SO_4 + 2CO_2 + 2H_2O

Volume of H_2SO_4 = 28 mL (given)

Since, 1 L = 1000 mL

So, 28 mL = 0.028 L

Molarity of H_2SO_4 = 5.4 M (given)

Molarity = \frac{number of moles of solute}{Volume of solution in Liters}   -(1)

Substituting the values in equation (1):

5.4 mol/L = \frac{number of moles of H_2SO_4}{0.028 L}

number of moles of H_2SO_4 = 5.4 mol/L\times0.028 L

number of moles of H_2SO_4 = 0.1512 mol

From the balanced reaction between H_2SO_4 and NaHCO_3, 2 moles of NaHCO_3 reacts with 1 mole of H_2SO_4.

Molar mass of NaHCO_3 = 84.007 g/mol

Mass of NaHCO_3 needed:

0.1512 mol H_2SO_4 \times \frac{2 mole NaHCO_3}{1 mole H_2SO_4}\times \frac{84.007 g}{1 mole NaHCO_3} = 25.404 g

Hence, the required amount of NaHCO_3 is 25.404 g.


7 0
3 years ago
25C and 1.0 atmosphere
adelina 88 [10]

Answer:

23

Explanation:

4 0
3 years ago
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