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uranmaximum [27]
3 years ago
5

The ability for an pond to freeze over in the winter

Chemistry
2 answers:
Otrada [13]3 years ago
8 0

Answer:

The ability for an pond to freeze over in the winter

Explanation:

The ability for an pond to freeze over in the winter

Brums [2.3K]3 years ago
3 0

Answer:

The ability for a pond to freeze over in the winter is a physical property.

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What intermolecular force involves non polar molecules
Zolol [24]

Dispersion forces are the only type of intermolecular force operating between non-polar molecules, for example, dispersion forces operate between hydrogen (H2) molecules, chlorine (Cl2) molecules, carbon dioxide (CO2) molecules, nitrogen tetroxide (N2O4) molecules and methane (CH4) molecules.

www.ausetute.com.au/intermof.html


7 0
3 years ago
The element silver exists in nature as two isotopes: 107Ag has a mass of 106.9051 u, and 109Ag has a mass of 108.9048 u. The ave
Lilit [14]

Answer:

107Ag has abundance of 51.7%

109Ag has abundance of 48.3%

Explanation: Please see attachment for explanation

3 0
3 years ago
If the density of water is 1g/ml, what is the mass of 254ml of water?
Mariulka [41]
The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer.  Since mL is in the numerator and denominator, mL cancels out and we are left with g only. 
8 0
3 years ago
How many grams in 6.20 x 10^25 atoms of bromine (Br) ? image attached , will give brainliest
Deffense [45]

Answer:

8239.2g

Explanation:

Given parameters:

Number of atoms in Br  = 6.2 x 10²⁵atoms

Unknown:

Mass of Br = ?

Solution:

From mole concepts, we know that:

       1 mole of a substance contains 6.02 x 10²³ atoms/mol

 Molar mass of Br  = 80g/mol

6.2 x 10²⁵atoms  x \frac{1}{6.02 x 10^{23} } \frac{mol}{atoms} x  80 x \frac{g}{moles}  

          = 8239.2g

8 0
3 years ago
Carbon dioxide (CO2) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "s
Lunna [17]

Answer:

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

Step 1: Given data

The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)

The sample hasa volume of 25.0 mL

Step 2: Calculating mass of the sample

The density is the mass per amount of volume

0.469g/cm³ = 0.469g/ml

The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

5 0
3 years ago
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