If a solution at pH 5 undergoes a 1000-fold increase in [OH-], what is the resulting pH?

heated blocks of iron used to be warm bed. A 1600g brick of iron heated to 155c would release how many joules of heat energy as

omeli [17]

We are going to use this formula :

Q = M*C*ΔT

when Q is the heat released by the material in J

and M is the mass in gram = 1600 g

C is the specific heat capacity of iron = 0.4494 J/g

and ΔT is the changing in temperature = 155 - 25 = 130 °C

So by substitution:

∴ Q = 1600 g * 0.4494 J/g * 130 °C

= 93475 J

∴ The amount of heat released in Joule = 93475 J

3 0

2 years ago

6.

RUDIKE [14]

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

3 0

2 years ago

A silver sphere has a mass of 5.492 g and a diameter of 10.0 mm. What is the density of silver metal in grams per cubic centimet

Blizzard [7]

Answer:

Explanat $\rho=10.5\frac{g}{cm^3}$ ion:

Hello,

In this case, considering the given diameter which is related to a radius of 5.0 mm and the formula for the calculation of the volume of the sphere, its volume in cubic centimeters (5.00 mm = 0.5 cm) is then:

$V=\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*(0.5cm)^3\\\\V=0.524cm^3$

In such a way, the density turns out:

$\rho =\frac{m}{V} =\frac{5.492g}{0.048m^3} \\\\\rho=10.5\frac{g}{cm^3}$

Best regards.

8 0

1 year ago

Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta

Arlecino [84]

Answer:

A. $\lambda_0=2.196\times 10^{-7}\ m$

Explanation:

The work function of the Platinum = $9.05\times 10^{-19}\ J$

For maximum wavelength, the light must have energy equal to the work function. So,

$\psi _0=\frac {h\times c}{\lambda_0}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda_0$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

Thus,

$9.05\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\frac{9.05}{10^{19}}=\frac{19.878}{10^{26}\lambda_0}$

$9.05\times \:10^{26}\lambda_0=1.9878\times 10^{20}$

$\lambda_0=2.196\times 10^{-7}\ m$

8 0

1 year ago

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a

geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0

2 years ago