Which of the following could not happen to an igneous rock?

How many grams are there in 1.18 X 1026 atoms of iridium?

stealth61 [152]

You want to divide by avagadros number (6.22 x 10^23). This will cancel the atoms unit and give moles, moles of Iridium. Now you want to calculate the atomic mass of Iridium which is in units of grams per mole. Multiply these two numbers and the moles will cancel giving you grams.

Setting up a dimension analysis type of thing helps tremendously. See what you have to cancel in order to get what you want. We canceled the atoms, then we canceled the moles, and then we got grams.

4 0

3 years ago

Humanities I

Inessa05 [86]

Answer:

✓ scholastics

Explanation:

you d.ont need a expla.nation rig.ht un.less y.ou wan.na re.ad for an h.our

3 0

2 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

Worldwide, the single most important chemical weathering agent is ________. worldwide, the single most important chemical weathe

Juliette [100K]

The single most important chemical weathering agent is Carbon dioxide.
Weathering refers to the process that change the physical and chemical character of rock at or near the surface. Weathering has a dramatic impact on the composition of Earth's atmosphere. Chemical weathering removes carbon dioxide from the atmosphere, allowing it to be transformed into limestone and stored in the crust. Without chemical weathering, the elevated levels of carbon dioxide in the atmosphere would have long made Earth too hot to sustain life.

4 0

3 years ago

Sodium hydroxide, NaOH, is a strong base that is used in industrial synthesis and processes such as making paper.

geniusboy [140]

No' of molecules divide by avogadro number , 6×6.023×10^23 so (2.2×10^22)÷(6.023×10^23)
= 0.03653 moles
moles × Molar mass = mass
n×Mr=m
0.03653×40 = 1.46 grams

8 0

3 years ago

Read 2 more answers