Which character of Life distinguishes a fire and a flowering plant

A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final

salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of $HNO_3$ = 13.5 M

$M_2$ = concentration of diluted solution = ?

$V_1$ = volume of $HNO_3$ = 25.0 ml = 0.0250 L

conversion used : (1 L = 1000 mL)

$V_2$ = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

$13.5M\times 0.0250L=M_2\times 0.500L$

$M_2=0.675M$

Therefore, the concentration of the diluted solution is 0.675 M

8 0

2 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

Describe how the surface area affect the rate of a reaction

leva [86]

When we increase the surface area of an object, more atoms are exposed. Since more atoms are exposed, the atoms can react faster, and this is why the rate of a reaction increases when the surface area increases.

For example, lets say we want to heat a potato. If we just put the whole potato in the microwave, it will take a long time for the potato to get thoroughly heated. However, if we chop the potato into smaller pieces, we will observe that it gets heated much faster. This is because we increased the surface area of the potato, which resulted in more potato atoms to be exposed to the heat, and caused the reaction to be faster.

7 0

2 years ago

I’ll give brainliest please help thanks.

oksano4ka [1.4K]

Answer:

See Explanation

Explanation:

The equation of the reaction is;

C7H16(g) + 11O2(g) ---->7CO2(g) + 8H2O(g)

1a) Number of moles in 228 g of H2O = 228 g/18 g/mol = 12.67 moles

From the reaction equation;

11 moles of O2 yields 8 moles of H2O

x moles of O2 yields 12.67 moles of H2O

x = 11 * 12.67/8

x = 17.4 moles of O2

Since 1 mole of O2 occupies 22.4 L

17.4 moles of O2 occupies 17.4 moles * 22.4 L/1 mole = 389.76 L

1b) Molar mass of C7H16 = 100 g/mol

Number of moles in 300 g of C7H16 = 300 g/100g/mol = 3 moles

1 mole of C7H16 yields 7 moles of CO2

3 moles of C7H16 yields 3 * 7/1 = 21 moles of CO2

If 1 mole of CO2 occupies 22.4 L

21 moles of CO2 occupies 21 moles * 22.4 L/1 mole = 470.4 L of CO2

2) Number of moles in 202 L of H2 is obtained by;

1 mole of H2 occupies 22.4 L

x moles of H2 occupies 202 L

x = 1 mole * 202 L/22.4 L = 9 moles of H2

From the reaction equation;

3 moles of H2 yields 2 moles of PH3

9 moles of H2 will yield 9 * 2/3 = 6 moles of PH3

Mass of 6 moles of PH3 = 6 moles of PH3 * 34 g/mol = 204 g of PH3

3 0

2 years ago

How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?

Nina [5.8K]

11.48-gram of $KCl0_3$ are needed to produce 6.75 Liters of $O_2$ gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$