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Lemur [1.5K]
3 years ago
8

Iron filings could be separated from sand using _______.

Chemistry
1 answer:
Deffense [45]3 years ago
3 0

Answer:

A. a magnet

Explanation:

Iron fillings could be separated from sand using a magnet

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If an object loses electrons it will become negatively charged. <br><br><br> True<br><br> False
Ray Of Light [21]
The answer should be true.
3 0
3 years ago
Which of these four elements is the most reactive metal?​
Ksivusya [100]

Answer:

Sodium (NA)

Explanation:

5 0
3 years ago
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Measure the diameter of the circle using Ruler A and Ruler B.Given that the actual diameter of the circle is 2.264 cmcm, classif
leva [86]

Answer:

<u>Ruler A :</u>

According to ruler A, the diameter of the circle has only one certain digit and one uncertain digit. if we look at measurement of a diameter, it contain two significant figures. so the certainty  of the diameter measurement is <em>smaller. </em>

<u>Ruler B :</u>

According to ruler B, the certainty  of the diameter measurement is <em>greater. </em>because it contain two certain and one uncertain digit. it has three significant figures.

5 0
3 years ago
How many grams of steam at 100 °C would be required to raise the temperature
Alex777 [14]

The mass of steam required to raise the temperature of water is 3.5 g.

The given parameters;

  • <em>mass of the benzene, = 47.6</em>
  • <em>initial temperature of the benzene, = 5.5 ⁰C</em>
  • <em>final temperature of the benzene = 30 ⁰C</em>

The molar mass of Benzene = 78.11 g/mol

The molar mass of water = 18 g/mol

The number of moles of the Benzene is calculated as follows;

n = \frac{47.6}{78.11} = 0.61 \ mole

The mass of steam required is calculated as follows;

<em>heat lost by steam = heat absorbed by benzene</em>

<em />

\frac{m}{18} \times 40.7 \times 10^3 = 47.6(1.63)(30-5.5) \ + \ 0.61 \times 9.87 \times 10^3\\\\2261.11 m = 7921.61\\\\m = \frac{7921.61}{2261.11} \\\\m = 3.5 \ g

Thus, the mass of steam required to raise the temperature of water is 3.5 g.

Learn more here:brainly.com/question/14963365

3 0
3 years ago
A piece of gold wire has a diameter of .175cm. How much will precisely 1.00 x 10^5 cm of the wire weigh
Alona [7]
The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be

V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³

Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire
7 0
3 years ago
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