All categories

3 years ago

Please help!

Chemistry

1 answer:

lyudmila [28]3 years ago

4 0

<h3>The equation</h3>

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, ie the heat received/absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

A calorimeter is a device used to measure the specific heat of material

A metal is put into a calorimeter that contains water and there will be heat transfer:

$\tt \displaystyle m_mc_m (T_m-T)=m_wc_w(T-Tw)$

m = metal

w = water

T = the final temperature of the mixture

mass of metal =(Nickel) = 4.94 g

mass of calorimeter = 12.5 g

mass of water = 123.13 g (135.63 - 12.5)

The equation

Q released (metal) = Q absorbed(calorimeter+water)

Qmetal = 4.94 x c . (Tm-23.60)
Q calorimeter = C.(23-6-21.9) --> C = heat capacity of calorimeter
Q water = 123.13 x c. (23.6-21.9)

The equation :

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

You might be interested in

How many grams are in 1.76 x 10^23 atoms of iodine

Mariana [72]

Answer:

$\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}$

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

$6.022*10^{23}$

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

$\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Multiply the given number of atoms by Avogadro's number.

$1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Flip the fraction so the atoms of iodine will cancel out.

$1.76*10^{23} \ atoms \ I*\frac{ 1 \ mol \ I}{6.022*10^{23} \ atoms \ I}$

$1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }$

Multiply so the problem condenses into 1 fraction.

$\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }$

$0.2922617071 \ mol \ I$

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

$\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply this fraction by the number of moles found above.

$0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply. The moles of iodine will cancel.

$0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }$

The 1 as a denominator is insignificant.

$0.2922617071 *{ 126.9045 \ g\ I }$

$37.08932581 \ g \ I$

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

$37.08932581 \ g \ I$

The 8 in the hundredth place tells us to round the 0 up to a 1.

$\approx 37.1\ g \ I$

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0

3 years ago

Do any of the spheres exist on their own or are they intertwined?

Sauron [17]

Answer:

show me a pic of the problem

Explanation:

show me a pic of the problem

3 0

3 years ago

What can you see in both the Map View and the Cross-Section View?

wlad13 [49]

Geologic Illustrations. A map shows a bird's eye view looking down at the earth's surface. ... Often a geologic map will indicate both the type of rock and the age of the rock. A cross section shows the surface of the earth (with or without topography or elevation) and a vertical slice of the earth

7 0

3 years ago

Read 2 more answers

Nitrogen monoxide, NO, reacts with hydrogen, H₂, according to the following equation.

Anvisha [2.4K]

Answer : The rate law for the overall reaction is, $Rate=k[NO]^2[H_2]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $2NO+H_2\rightarrow N_2+H_2O_2$ (slow)