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meriva
3 years ago
11

Which is an electronic tool that can be used to improve science?

Chemistry
2 answers:
Neporo4naja [7]3 years ago
8 0
A beaker probeware is the answer I think
Vanyuwa [196]3 years ago
6 0
The answer is probeware
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Earth’s inner core:_____ as Earth’s outer core:liquid
KatRina [158]

Answer:However, unlike the outer core, the inner core is not liquid or even molten. The inner core's intense pressure—the entire rest of the planet and its atmosphere—prevents the iron from melting. The pressure and density are simply too great for the iron atoms to move into a liquid state.

Explanation:please give brainliest

5 0
2 years ago
05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises
OLga [1]

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

6 0
3 years ago
Jessica jogs on a path that is 25 km long to get to a park that is south of the jogging path if it takes us a good 2.5 hours the
zysi [14]
She jogs 10km per hour. Hope this helps :)
3 0
3 years ago
Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)
Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope: </u>

% = 50.69 %

Mass = 78.9183 amu

<u>For second isotope: </u>

% = 49.31 %

Mass = 80.9163 amu

Thus,  

Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu

Average\ atomic\ mass=40.0036+39.8998\ amu

<u>Average atomic mass = 79.9034 amu</u>

4 0
3 years ago
Hey pleaseeeeeeeeeeeeeeeeeeeeeeee
charle [14.2K]

Answer: B

Explanation:

6 0
3 years ago
Read 2 more answers
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