Your answer should have the same number of significant figures as the starting measurement.

I need help with balancing equations I'm doing homework and idk what to do here can you give me the answers please lol

amm1812

It's pretty easy to balance equations! Basically you want to make sure that the number of each compound is equal on both sides of the arrow.

For example number one is

Fe + H2SO4 -> Fe2(SO4)3 + H2

A 3 in front of H2SO4 because there's a subscript of 3 on the right side.

Then a 3 in front of H2 because of the previous step.

Then add a 2 in front of Fe because of the 2 subscript in Fe2(SO4)3

Then add a 1 in front of Fe2(SO4)3 because you already have an equal number of each element.

2Fe + 3H2SO4 -> 1Fe2(SO4)3 + 3H2

I hope this explanation helps! You should really do your homework because practice is everything when it comes to chemistry. You'll need to know how to do it for exams.

3 0

3 years ago

Please begging you guys someone help me

BaLLatris [955]

Answer:

See Explanation

Explanation:

10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.

11) The bombardment of U-235 with neutrons leads to the reaction;

$U\frac{235}{92} + n\frac{1}{0}---> I\frac{138}{53} + Y\frac{95}{39} +3n \frac{1}{0}$

Hence

a = 92, b= 95, c= 53

12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.

Hence;

$Th\frac{231}{90} -----> e\frac{1}{0} +Ac \frac{231}{89}$

4 0

3 years ago

Calculate the change in energy when 75.0 grams of water drops from 31.0C to 21.6.

zysi [14]

Answer: Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

5 0

3 years ago

Read 2 more answers

What type of energy transformation is taking place when natural gas is used to heat water

kifflom [539]

That is thermal energy

8 0

3 years ago

The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ

vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

$q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m} =125.28W$

The heat flux can be defined as the amount of heat flow by unit of area.

Using the previous calculation, we can estimate the heat flux:

$heat \, flux=\frac{q}{A}=\frac{125.28 W}{9 m^{2} } =13.92 W/m^{2}$

It can also be calculated as:

$q/A=-k*\frac{\Delta T}{\Delta X}$

The thermal resistance can be expressed as

$\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}$

For the 3m x 3m sheet, the thermal resistance is

$R_t = \frac{\Delta X}{k*A}=\frac{0.025m}{0.029W/mK*9m^{2}}=0.0958 \, K/W$

4 0

3 years ago