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2 years ago

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA HELP

Chemistry

2 answers:

skad [1K]2 years ago

7 0

Answer:

Explanation:

mark me brainliest pls

Naya [18.7K]2 years ago

5 0

Coco beans poops okkkkk

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R.I.P too my brainly account lol..

Vera_Pavlovna [14]

Answer:

i don't know what happened but r.i.p account

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3 years ago

Read 2 more answers

Ionic crystals are excellent insulators and can hold a large amount of heat before melting or boiling.

sweet-ann [11.9K]

It is a true fact that ionic crystals are excellent insulators and can hold a large amount of heat before melting or boiling. The correct option among the two options that are given in the question is the first option. Salt is a great example of ionic crystals and we know that it takes a huge amount of time to melt or boil.

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3 years ago

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.

maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.

Explanation:

The molecular formula of boron trifluoride is $BF_3$ .

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- $6.023x10^2^3$ atoms.

Then, 2.20 moles of boron has

- $=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms$

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

$(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol$

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0

2 years ago

Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will r

Alexandra [31]

To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

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3 years ago

What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3

Dmitriy789 [7]

$d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\
V=4,2cm^{3}\\\\
m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g$

6 0

3 years ago