What is the oxidation state of nitrogen in NaNO2? (1) +1 (3) +3 (2) +2 (4) +4

4. By what quantity must the heat capacity be divided to obtain the specific heat of that material? (I poin

charle [14.2K]

Answer:

its mass

its temperature

Explanation:

The heat capacity of a body is the amount of heat needed to raise that temperature of a body by 1k or 1°C.

It is expresses as:

H = m c Ф

where m is the mass of the body

c is the specific heat of the body

Ф is the temperature change of the body

Specific heat capacity is that heat content needed to raise the temperature of a unit mass of a substance by 1k or 1°C

Heat capacity is also known as heat content;

To solve the problem:

since H = m c Ф

we make the unknown c the subject of the expression;

c = $\frac{H}{m (t_{2} - t_{1} )}$ = H / mФ

So to obtain the specific heat capacity, divided the heat capacity by mass and its temperature change.

8 0

3 years ago

Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip

erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0

3 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

A barium hydroxide solution is prepared by dissolving 2.06 g of Ba(OH)2 in water to make 32.9 mL of solution. What is the concen

navik [9.2K]

Answer: 1.36 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

To calculate the moles, we use the equation:

moles of solute= $\frac{\text {given mass}}{\text {molar mass}}=\frac{2.06g}{171g/mol}=0.0120moles$

$Molarity=\frac{0.0120\times 1000}{32.9}=0.364M$

The balanced reaction between barium hydroxide and perchloric acid:

$2HCIO_4+Ba(OH_)2\rightarrow BaCIO_4+2H_2O$

To calculate the concentration of acid, 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 $HClO_4$

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

We are given:

$n_1=1\\M_1=?\\V_1=8.50mL\\n_2=2\\M_2=0.364M\\V_2=15.9mL$

Putting values in above equation, we get:

$1\times M_1\times 8.50=2\times 0.364\times 15.9\\\\M_1=1.36M$

Thus the concentration of the acid is 1.36 M

4 0

3 years ago

You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col

miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0

3 years ago