If an electron is released during radioactive decay which type of Decay has taken place

A hot gas flowing through a pipeline can be considered as a:________

k0ka [10]

Answer:

B) irreversible process

Explanation:

The process given here is irreversible.

7 0

3 years ago

A marble rolls off horizontally from the edge of table top 1.50 m above the floor. it strikes the floor 2.0 m from the base of t

GrogVix [38]

a. t=0.553 s

b. vox(horizontal speed) = 3.62 m/s

<h3>Further explanation</h3>

Given

h = 1.5 m

x = 2 m

Required

a. time

b. vo=initial speed

Solution

Free fall motion

a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)

$\tt t=\sqrt{\dfrac{2h}{g} }$

t = √2h/g

t = √2.1.5/9.8

t=0.553 s

b. x=vox.t(horizontal motion)

$\tt x=v_{ox}\times t$

vox=x/t

vox=2/0.553

vox=3.62 m/s

3 0

2 years ago

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago

1.24 moles of magnesium arsenate are dissolved in 1.74 kg of solution. Calculate the molality of the solution.

aleksklad [387]

Answer:

Molality of the solution = 0.7294 M

Explanation:

Given:

Number of magnesium arsenate = 1.24 moles

Mass of solution = 1.74 kg

Find:

Molality of the solution

Computation:

Molality of the solution = Mole of solute / Mass of solution = 1.74 kg

Molality of the solution = 1.24 / 1.7

Molality of the solution = 0.7294 M

6 0

3 years ago

05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises

OLga [1]

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

6 0

3 years ago