How many grams of H20 are created from 1.3 liters of H2?

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio

koban [17]

Answer:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

Explanation:

In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case $Pt^+^2$ ) and the anion is the atom at the right of the name (in this case $Cl^-$ ). With this in mind, the formula would be $PtCl_2$ .

Now, if we used metallic copper we have to put in the reaction only the copper atom symbol $Cu$ . So, we have as reagents:

$Cu~+~PtCl_2->$

The question now is: What would be the products? To answer this, we have to remember "single displacement reactions". With a general reaction:

$A~+~BC->AB~+~C$

With this in mind, the reaction would be:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

I hope it helps!

5 0

3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

Which of the following has helped preserve resources while cutting the demand for energy?

sweet [91]

A is the correct answer

4 0

3 years ago

Read 2 more answers

A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c

Kamila [148]

We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:

KOH + H₂SO₄ → H₂O + KHSO₄

If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:

0.025 L x 0.150 mol/L = .00375 mol KOH

0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄

We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:

0.00375 mol / 0.015 L = 0.25 mol/L

The concentration of H₂SO₄ being neutralized is 0.25 M.

6 0

3 years ago