CHEMISTRY HIGH SCHOOL

A sample of seawater has a mass of 158g and a volume of 156mL. What is its density?

Answers

Answer 1

Answer:

A sample of seawater has a mass of 158g and a volume of 156mL. its density is 1.012 g/mL.

According to the question the data are given is as follows:

volume of sea water = 156 mL

mass of sea water = 158 g

density of sea water can be calculated by the following formula :

D = M / V

where,

D = density

M = mass of sample

V = volume of sample

substituting all the value in the given formula , we get :

D = 158 g / 156 mL = 1.012 g/mL

Density of seawater = 1.012 g / mL

Thus, A sample of seawater has a mass of 158g and a volume of 156mL. its density is 1.012 g/mL .

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3.54 mL

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = (mass)/(density) \n$

From the question

mass = 9.38 g

density = 2.65 g/cm³

The volume of the quartz is

$volume = (9.38)/(2.65) \n = 3.539622641...$

We have the final answer as

3.54 mL

Hope this helps you

Which is the safest method for diluting concentrated sulfuric acid with water?

Answers

The safest method for diluting concentrated sulfuric acid with water is to add acid to water. This way, when spill occurs, the acid is already diluted and less harmful than adding water to acid.

What subatomic particle(s) is/are located in the nucleus of the atom? Protons only

Electrons only

Protons and neutrons

Protons and electrons

Answers

Protons and neutrons!

Give the number of significant figures: 3.10 cm

Answers

There are 3 significant figures.

Non-zero digits are always significant.

Any zeros between two significant digits are significant.

A final zero or trailing zeros in the decimal portion ONLY are significant.

You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2.00 *10^4 Kpa at 28c How many kilograms of N2 does the cylinder contain

Answers

The quantity of nitrogen the cylinder contains is 4477.8 g

What is pressure?

Pressure is a force exerted in a perpendicular direction in any item.

By ideal gas law

PV = nRT

$P =(w)/(M) =(RT)/(V)\n$

w = mass

Volume is 20.0 l

Pressure is $2.00 * 10^4 \;Kpa$

The molar mass of nitrogen is 28 g/mol

R is gas constant = 0.0821

Temperature is 28 converted into kelvin that is 301 k

Putting the values

$197.6 =(w)/(28) =(0.0821 * 301 )/(20\;l)\n\nw = 4477.8 g$

Thus, the mass of nitrogen is 4477.8 g.

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Answer : The mass of $N_2$ gas is, 4477.8 g

Solution :

using ideal gas equation,

$PV=nRT\n\nP=(w)/(M)* (RT)/(V)$

where,

n = number of moles of gas

w = mass of gas

P = pressure of the gas = $2* 10^4Kpa=197.6atm$

conversion : $1atm=101.2Kpa$

T = temperature of the gas = $28^oC=273+28=301K$

M = molar mass of $N_2$ gas = 28 g/mole

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 20 L

Now put all the given values in the above equation, we get the mass of gas.

$197.6atm=(w)/(28g/mole)* (0.0821Latm/moleK* 301K)/(20L)$

$w=4477.8g$

Therefore, the mass of $N_2$ gas is, 4477.8 g

Consider the following balanced chemical equation. 4 KO 2 + 2 H 2 O ⟶ 4 KOH + 3 O 2 How is the rate of appearance of O 2 , Δ [ O 2 ] Δ t , related to the rate of disappearance of KO 2 ?

Answers

The appearance of Oxygen molecule is directly related or dependent to the rate of disappearance of potassium superoxide (KO2).

The rate of appearance of O2 is directly related to the rate of disappearance of KO2 because potassium superoxide (KO2) has oxygen molecule which is attached to the potassium so when the potassium reacts with water molecules, the potassium hydroxide (KOH ) and oxygen molecules is formed.

If the chemical reaction occurs then we get oxygen molecules so we can conclude that appearance of Oxygen molecule is directly related or dependent to the rate of disappearance of potassium superoxide (KO2).

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Answer : The rate of appearance of $O_2$ related to the rate of disappearance of $KO_2$ is,

$(d[O_2])/(dt)=-(3)/(4)(d[KO_2])/(dt)$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$4KO_2+2H_2O\rightarrow 4KOH+3O_2$