Which is the average kinetic energy of particles in an object?

How many moles of MgSiO3 are in 237g of the compound?

Anna35 [415]

For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16

Mr = 24 + 28 + 16 x 3
Mr = 100

moles = mass / Mr
moles = 237 / 100
moles = 2.37

6 0

3 years ago

Do subatomic particles have mass

katovenus [111]

Answer:

Yes

Explanation:

Masses for the three subatomic particles can be expressed in amu (atomic mass units) or grams. For simplicity, we will use the amu unit for the three subatomics. Both neutrons and protons are assigned as having masses of 1 amu each.

5 0

3 years ago

What does the atomic mass of an element on the periodic table represent?

Lynna [10]

The atomic mass on the periodic table represents the sum of number of protons and number of neutrons.

Atomic mass = Number of protons + number of neutrons

Hope this helps!

3 0

2 years ago

The atom as a whole<br> doesn’t carry any electric charge. It’s called a ____________ _________.

olchik [2.2K]

Answer:

Neutron has no charge while electron has a negative charge and proton has a positive charge

5 0

3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

3 years ago