What is the molar mass for CaWO4

What would make bromine, a group 17 element, more similar to a noble gas?

nlexa [21]

Thank you so much men nice well

5 0

3 years ago

When dinitrogen pentaoxide, a white solid, is heated, it decomposes to produce nitrogen dioxide gas and

AysviL [449]

9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

Oxygen produced - 1.618 gram

Decomposition of $N_2O_5$ takes place.

Find - Amount of $NO_2$ produced.

The decomposition reaction is as follows -

$2N_2O_5$ --> $4NO_2 + O_2$

Moles of $O_2$ gas = $\frac{1.6}{16}$ =0.1 moles.

1 mole of $O_2$ is produced from 2 moles of dinitrogen pentoxide

0.1 mole of $O_2$ will be produced from = 0.2 moles.

Now, 2 moles of dinitrogen pentoxide produce 4 moles of $NO_2$

$NO_2$ produced will be - 0.4 moles.

Weight of $NO_2$ produced - 0.4 X 46

Weight of $NO_2$ produced - 18.4 gram

Thus, grams of $NO_2$ produced are 18.4

Now calculate the volume of $NO_2$

Given data are:

P=103.25 kPa =1.01899827 atm

T= 22.75 °C +273 = 295.75 K

n=0.4 moles

V=?

R= 0.0821 liter·atm/mol·K

Putting the value in PV=nRT

V = $\frac{nRT}{P}$

V = $\frac{0.4 \;moles \;X \;0.0821\; liter\;atm/\;mol \;K X \;295.75 \;K}{1.01899827 atm}$

V= 9.5314 L

Hence, 9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

Learn more about the ideal gas equation here:

brainly.com/question/13450124

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8 0

2 years ago

How many moles are in 8.90 X 10^24 atoms of Na?

mixas84 [53]

Answer:

1.48 x 10^47 mol of Na

Explanation:

8.90 x 10^24 atoms of Na (1 mol of Na/6.022 x 10^23 atoms of Na)=

1.48 x 10^47 mol of Na

5 0

3 years ago

Imagine that you have an ideal gas in a 6.70 L container, and that 1450 molecules of this gas collide with a square-inch area of

kakasveta [241]

Answer:

they could collide 241.66 molec / in² by increasing the volume to 40.2L

Explanation:

ideal gas:

<u>Boyle Law</u>: at constant temperature the pressure of a gas varies inversely with the volume

V1 * P1 = V2 * P2
P = F / A

∴ V1 = 6.70 L;

∴ P1 = 1450 molec / in²

∴ V2 = 40.2 L

⇒ P2 = (( 6.70 L ) * ( 1450 molec/in²)) / 40.2 L

⇒ P2 = 241.66 molec/in²

6 0

3 years ago

Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Ide

olya-2409 [2.1K]

Answer:

strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

$X^{+}$ will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

$E_{X^{+}\mid X}^{0}(1.52V)> E_{Z^{3+}\mid Z}^{0}(0.84V)> E_{Y^{2+}\mid Y}^{0}(-1.17V)$

So, strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

As reduction potential of the half cell $X^{+}\mid X$ is higher than the reduction potential of the half cell $Z^{3+}\mid Z$ therefore $X^{+}$ will oxidize Z into $Z^{3+}$ and itself gets converted into X.

5 0

3 years ago