How much heat is required to raise the temperature of 25.0 grams of waterfrom 75.0°C to its boiling point?

A gas occupies a volume of 4.50 L at 44 kPa. What would be the new volume at 50 kPa?

oksano4ka [1.4K]

Explanation:

P1= 44 kpa

P2= 50 kpa

V1= 4.50 L

V2= ?

P1 V1= P2 V2

44 × 4.50 = 50 × V2

198= 50 × V2

V2 = 198/ 50

V2= 3.96 L "the new volume"

5 0

3 years ago

What is heat energy in science

Ket [755]

Answer:

Heat energy is the result of the movement of tiny particles called atoms, molecules or ions in solids, liquids and gases. Heat energy can be transferred from one object to another. the transfer or flow from one object to another is called heat.

hopefully this helped :3

4 0

3 years ago

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper

Olin [163]

Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

= (M)(Volume of the solution (L))

= (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

= Qrxn/(no. of moles of AgNO₃)

= (1170.4 J)/(0.03 mol)

= 39013.33 J/mol

= 39.013 kJ/mol.

7 0

3 years ago

Is zn2+ a transition metal?

4vir4ik [10]

Answer:

Yes.

Explanation:

Zn2+ is the zinc ion in aqueous solution.

Zinc is a transition metal.

These metals are in the middle of the periodic table and have similar properties.

7 0

3 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago