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Alecsey [184]
3 years ago
7

Which element can join with other elements to form covalent bonds? beryllium (Be) carbon (C) cesium (Cs) iron (Fe)

Chemistry
2 answers:
Agata [3.3K]3 years ago
6 0

Answer:

It’s C

Explanation:

Rainbow [258]3 years ago
4 0

Answer:

Carbon (C)

Explanation:

Carbon is the only nonmetal there and covalent bonds happen between two nonmetal atoms

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Is sublimation exothermic or endothermic
Masja [62]

Answer:

endormic

Explanation:

It occurs at a temperature and pressures below a substance's triple point on its phase diagram, which corresponds to the lowest pressure at which the substance can exist as a liquid

5 0
3 years ago
How many liters of C3H6O are present in a sample weighing 25.6 grams?
Romashka [77]

Answer:

V = 0.0327 L.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the liters of C3H6O by the definition of density. We can tell the density of this substance as that of acetone (0.784 g/mL) and therefore calculate the liters as shown below:

V=25.6g*\frac{1mL}{0.784g}*\frac{1L}{1000mL}\\\\V=0.0327L

Regards!

7 0
3 years ago
How do I round 10.25 to only 3 significant digits
Lady_Fox [76]
10.3 is good the correct rounding of three sig figures

3 0
4 years ago
A pure white crystalline compound was found to melt at 112.5-113.0oC when taken on a melting point apparatus, and on further hea
Sergeeva-Olga [200]

According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than  112°C is caused by reusing samples leads to errors.

A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).

However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above  112°C.

Learn more: brainly.com/question/5325004

8 0
3 years ago
A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calc
disa [49]

Answer:

specific heat = 0.951 j/g·°C

Explanation:

Heat flow equation  => q = m·c·ΔT

q = heat flow = 4817 joules

m = mass in grams = 140 grams Aluminum

c = specific heat = ?

ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C

q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C

6 0
3 years ago
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