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3 years ago

HELP QUICKLY PLEASE!!! Thermal energy is transferred from a hot object to a colder object by means of ___

Chemistry

1 answer:

Butoxors [25]3 years ago

5 0

Answer:

Heat is the transfer of energy

Explanation:

Heat refers to the energy transferred from a hotter object to a cooler one. Heat is transferred in three ways: radiation, conduction, and convection. Radiation is the direct transfer of energy. The direct transfer of heat from one substance to another substance that it is touching is called conduction.

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VARVARA [1.3K]

Answer:

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Explanation:

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3 years ago

What is the density (g/ml) of a sample of mineral oil . if 250ml has a mass of 0.23kg

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3 years ago

Read 2 more answers

Calculate the energy released when 24.8 g Na2O reacts in the following reaction.

gizmo_the_mogwai [7]

Answer:

= 48.077 kcal

Explanation:

The balanced equation is :

$Na_{2}O(s)+2HI(g)\rightarrow 2NaI(s)+H_{2}O(l)$

$\Delta H=-120Kcal$

Since the stoichiometric coefficient of Na2O = 1 hence its heat is

-120 Kcal/mol rather -120 Kcal only.

This means when 1 mole of the Na2O is burned , then 120 kcal of energy is released .

Molar mass of Na2O = 2(mass of Na) + 1(mass of O)

= 2(22.98)+16

=61.97 gram/mol

Molar mass = mass of the substance in grams present in 1 mole of it.

In 1 mole of Na2O , 61.97 gram of it is present.

Hence

61.97 gram of Na2O = -120 Kcal of heat is released

1 gram of Na2O =

$/frac{-120kcal/mol}{61.9gram/mol}$

So,

24.8 gram of Na2O will produce,

$/frac{-120kcal/mol}{61.9gram/mol}/times 24.8$

= -48.077 kcal of heat

You can also proceed through mole concept:

Calculate number of moles in 24.8 gram Na2O

$Moles=\frac{given\ mass}{Molar\ mass}$

Moles of Na2O = 24.08/61.9 = 0.4007 moles

So these many moles will produces:

0.4007 x 120

=-48.07

7 0

3 years ago

Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following reaction equation: CH4 + 4 Cl2 → CCl4 + 4HCl. What is the Δ

Vlad1618 [11]

Answer: The ΔH of the reaction if 51.3 g of $CH_4$ reacts with excess $Cl_2$ to yield 1387.6 kJ is 432.27kJ

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $CH_4$

$\text{Number of moles of methane}=\frac{51.3g}{16g/mol}=3.21moles$

$CH_4+4Cl_2\rightarrow CCl_4+4HCl$ $\Delta H=?$

As $Cl_2$ is present in excess, $CH_4$ is the limiting reagent as it limits the formation of product.

If 3.21 moles of methane releases heat = 1387.6 kJ

Thus 1 mole of methane release= $\frac{1387.6}{3.21}\times 1=432.27kJ$

4 0

3 years ago

The combustion of a 0.4255 g sample of a compound containing carbon, hydrogen, oxygen and bromine produces 0.4961 g CO2 and 0.17

IgorC [24]

Answer: The empirical formula is C11HO3Br

Explanation: 1ST SAMPLE;

First, we need to get the number of moles;

Molar mass of C=12, O=16, H=1, Ag =107.9, Br=79.9

0.4961g CO2 x (1 mol CO2) / (44.0g CO2) = 0.0113 mol of Co2

Since one mole of CO2 is made up of 1 mole of C and 2 moles of O, and we have 0.0113 moles of CO2 in our sample, then we know we have 0.0113 moles of C in the sample. Now, we need to know the mass of C we have in the sample:

(0.0113 mol C) (12.011g C/1 mol C) = 0.136g C

Now we follow the same pattern above and do it for the hydrogen:

0.1777g H2O x (1 mol H2O) / (18g H2O) = 0.01 mol

Now, for the mass of H:

(0.01 mol H) (1g H/1 mol H) = 0.01g H

But this is for 1 mole of H. Whereas, H has 2 moles from the question, therefore it's equal to 0.01 x 2 = 0.02g H

Since we combusted 0.4255g of the sample, the missing mass will be from the bromine and oxygen since we have gotten masses of Carbon and Hydrogen.

Therefore missing mass = 0.4255 - (0.136+0.02) = 0.2695 of bromine and oxygen

2ND SAMPLE:

First, we need to get the number of moles;

0.1894 g AgBr x (1 mol AgBr) / (107.9 + 79.9g AgBr) = 0.001 mol of AgBr

Since AgBr is made up of 1 mole of Ag and 1 mole of Br each,

We can say that there's 0.001 mole of Br in this second sample.

Now looking at the first and second samples, lets set up a proportion to know the number of moles of Br in the first sample.

If 0.1523g of the second sample produced 0.001 mol of Br, therefore 0.4255g in the first sample will produce: (0.4255g x 0.001)/0.1523 = 0.0028mol of Br

Therefore the mass of Br in the first sample is: (0.0028 mol C) (79.9g Br/1 mol C) = 0.22372g Br

From the first sample, we saw that the sum of Br and Oxygen equals 0.2695

Therefore,the mass of oxygen is: 0.2695 - 0.22372 = 0.04578g of oxygen

Therefore to find number of moles of oxygen;

(0.04578g O x (1 mol O) / (16.0g O) = 0.0029 mol of oxygen

Overall, we have; C=0.0113 moles ;H=0.001 moles; Br= 0.0028moles and O= 0.0029

The smallest is 0.001. So to simplify this for the empirical formula, we divide each by 0.001 to get approximately C= 11, H=1, Br=3, O=3

Therefore the empirical formula is C11HO3Br

5 0

3 years ago