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3 years ago

If the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?

Chemistry

1 answer:

lana66690 [7]3 years ago

5 0

Answer:

2, strong acid

Explanation:

Data obtained from the question. This includes:

[H+] = 0.01 M

pH =?

pH of a solution can be obtained by using the following formula:

pH = –Log [H+]

pH = –Log 0.01

pH = 2

The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.

Since the pH of the above solution is 2, it means the solution is a strong acid.

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

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3 years ago

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A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is

Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V = π × 182.25 cm × 32.4 cm

V = 18550.79 cm³

V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

vessel contains 0.218kg or 218 gram of carbon monoxide gas

molar mass of carbon monoxide gas is 28.010 g/mol

moles of carbon monoxide gas n = 218 gram / 28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

T = ( 889.317387 - 273.15 ) °C

T = 616.167 ≈ 616 °C { 3 significant digits }

Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

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