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3 years ago

Which are precautions to take when working with heat and fire? Check all that apply.

1 answer:

3 0

Wear face mask and holy the conical flask or test tube away from your face and body and dont put it on directly on fire or it

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In scientific notation what is the radius of the sun

zloy xaker [14]

Answer:

4.32 × 10^5 miles

Explanation:

4 0

3 years ago

Carbon-12 contains 6 protons and 6 neutrons. The radius of the nucleus is approximately 2.7 fm (femtometers) and the radius of t

Slav-nsk [51]

Explanation:

Let us assume that volumes is equal to spheres volume of nucleus.

fm = femto meter = $10^{-15}$

pm = pico meter = $10^{-12}$

Hence, calculate the volume as follows.

V = $\frac{4}{3} \times \pi \times r^{3}$

= $\frac{4}{3} \times 3.14 \times (2.7)^{3}$

= 82.4481 $fm^{3}$ (cubic femtometers)

or, = 82 $fm^{3}$

Now, we will calculate the volume of the atom as follows.

V = $\frac{4}{3} \times \pi \times r^{3}$

= $\frac{4}{3} \times 3.14 \times (70)^{3}$

= 1436758.4 $pm^{3}$ (cubic picometers)

thus, we can conclude that volume of the nucleus is 82 $fm^{3}$ and volume of the atom is 1436758.4 $pm^{3}$ .

7 0

3 years ago

Ab cd and Ef intersect at p. If r=90, s=50, t=60, u= 45 and w=50, what is the value of x?

ziro4ka [17]

Answer:

Explanation:

3 0

3 years ago

Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the

Ira Lisetskai [31]

Answer: B aka #2

Explanation:

3 0

2 years ago

An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi

Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM → 54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0

3 years ago