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Firlakuza [10]
3 years ago
10

H2 how many elements?

Chemistry
1 answer:
rjkz [21]3 years ago
4 0
There are two elements, two hydrogen to be exact.
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Write the name for the compound whose formula is shown here<br><br> Ca3(BO3) 2
4vir4ik [10]

Answer:

Calcium Nitrate

Explanation:

3 0
3 years ago
Shelby measured the volume of a cylinder and determined it to be 54.5cm3. The teacher told her that she was 4.25% too high in he
FinnZ [79.3K]

Answer:

56.8cm³

Explanation:

Given parameters:

Measured volume = 54.5cm³

Percentage error  = 4.25%

Unknown:

Actual volume of the cylinder = ?

Solution:

The percentage error shows the amount of error introduced into a measurement.

We need to find out this amount of error from the data given.

   Error = Percentage error x measured volume

                       Error = \frac{4.25}{100} x 54.5  = ±2.32cm³

Since the error introduced = 2.32cm³

  Actual volume of cylinder = measured volume ± Error

  Since the percentage error quoted as higher 4.25%;

 Actual volume  = 54.5cm³ + 2.32cm³ = 56.8cm³

       

 

6 0
3 years ago
Which argument best explains the charge of an atomic nucleus
8_murik_8 [283]

Answer:

An atomic nucleus is positively charged because it is composed of protons.

Explanation:

hope this helps you:)

7 0
3 years ago
Read 2 more answers
What are the correct steps for evolution by natural selection?​
katrin [286]

(っ^-^ )っHey There!

Answer → The five steps of the process of natural selection are variation, inheritance, selection, time and adaptation

✨Hope this Helps✨

8 0
3 years ago
We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
3 years ago
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