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2 years ago

The temperature at which the reaction is occurring is increased

1 answer:

6 0

The addition of heat energy to a system always causes the temperature of that system to increase. This is always true because you are adding heat of a substance to increase its temperature. For example, you are going to drink a cup of coffee. And you wanted it hot to boost your attention. So you have to use hot water. In order for your water to become hot or warm, you need boil it in a kettle. Note that you are going to use an electric stove. The electric stove gets it energy from the source giving it a hotter temperature to the water in the kettle. You are applying heat energy to warm the water. So, the statement is true.

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AGO H2SO4 (aq) Al(SO4)3(aq) H28) fe Cl218) FeCl38<br>

Rina8888 [55]

Answer:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

2Fe + 3Cl2 → 2FeCl3

Explanation:

1. (SO4) 3 you see this 3 it means that 3 must be behind H2SO4. So now it's 3H2SO4.

2. If 3 is now behind one H2, it must be behind the other.

So now it's 3H2.

3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.

1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.

2. Since it is now 2FeCl3, we know that there must be 2 in the second Fe. It's 2Fe now.

7 0

2 years ago

What’s a tripping hazard in Science

attashe74 [19]

Spillages on a tiled floor, a highly polished floor surface, a damp smooth surface with no mat.

5 0

3 years ago

Read 2 more answers

An electron has an uncertainty in its position of 471 pm .

olya-2409 [2.1K]

<u>Answer:</u>

$\Delta x . \Delta p \geq \frac{h}{4 \pi}$

∆x is the Uncertainty in the position of the particle

∆p is the Uncertainty in the momentum of the particle

$\Delta p=\frac{h}{4 \pi \times m \times \Delta x}$

$=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} s^{-1}}{4 \times 3.14 \times 9.109 \times 10^{-31} \mathrm{kg} \times 471 \times 10^{-12} \mathrm{m}}$

$=\frac{6.626 \times 10^{-34} \mathrm{ms}^{-1}}{53887 \times 10^{-43}}$

$=1.23\times10^5 ms^{-1}$ is the answer

3 0

2 years ago

Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C

lisov135 [29]

The answer is 7345 g

5 0

3 years ago

When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.

sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $15.8J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m$ = mass of water = 100.0 g

$\Delta T$ = change in temperature = $T_2-T_1=(32.0-23.9)=8.1^oC$

Now put all the given values in the above formula, we get:

$q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]$

$q=3513.8J=3.5138kJ$ (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole$

Now,

$\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol$

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

4 0

3 years ago