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Sphinxa [80]
3 years ago
11

Long scratch marks that are left behind as evidence of glaciers are called

Chemistry
1 answer:
Archy [21]3 years ago
5 0

Answer:

it is A striations

Explanation:

..........

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The average propane cylinder for a residential grill holds approximately 18 kg of propane. how much energy (in kj) is released b
never [62]
In order to answer this, we mus know the data for the heat of combustion of propane. This is an empirical data that you can search online. The heat of combustion is -2220 kJ/mol. The molar mass of propane of 44.1 g/mol. The solution is as follows:

ΔH = -2220 kJ/mol (1 mol/44.1 g)(1000g/1kg)(20 kg)
<em>ΔH = -1006802.721 kJ or -1 GJ</em>
8 0
3 years ago
After three half-lives of an isotope, 1 billion (⅛) of the original isotope’s atoms still remain in a certain amount of this ele
erica [24]

You would expect 500 million atoms to be present. When a source experiences a half-life, it is a point in which atoms have degraded to half what they used to be in mass. 1/2(1 billion) = 500 million

Brainliest answer please

4 0
3 years ago
A lead ball is added to a graduated cylinder containing 41.7 ml of water, causing the level of the water to increase to 96.0 mL.
rjkz [21]

Answer:

Vlead ball=54.3 mL

Explanation:

The graduated cylinder contains 41.7mL of water

mL is a volume unit.

Water volume = 41.7 mL

The lead ball caused an increase of volume from 41.7 mL to 96.0 mL

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

96.0mL=Vleadball +41.7mL

Vlead ball=96.0mL-41.7mL

Vleadball=54.3mL

This is actually true if we suppose that the lead ball is fully sunken in the water.

We always must consider that the volume difference is the volume that the sunken object is occupying in the water.

5 0
3 years ago
What is the molecular geometry of COF2?
maxonik [38]

Answer:

trigonal planar

Explanation:

5 0
3 years ago
How many molecules of F2 react with 66.6 g NH3 ? (stoichiometry)
stiks02 [169]
<h3>Answer:</h3>

5.89 × 10^23 molecules of F₂

<h3>Explanation:</h3>

The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;

5F₂ + 2NH₃ → N₂F₄ + 6 HF

We are given 66.6 g NH₃

We are required to determine the number of fluorine molecules

<h3>Step 1: Moles of Ammonia </h3>

Moles = Mass ÷ Molar mass

Molar mass of ammonia = 17.031 g/mol

Moles of NH₃ = 66.6 g ÷ 17.031 g/mol

                      = 3.911 moles

<h3>Step 2: Moles of Fluorine </h3>

From the equation 5 moles of Fluorine reacts with 2 moles of ammonia

Therefore,

Moles of fluorine = Moles of Ammonia × 5/2

                            = 3.911 moles × 5/2

                           = 9.778 moles

<h3>Step 3: Number of molecules of fluorine </h3>

We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules

Therefore;

1 mole of F₂ = 6.022 × 10^23 molecules

Thus,

9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole

                              = 5.89 × 10^23 molecules

Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules

5 0
3 years ago
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