Pls help i am in 8th grade k12

You have used 2.4×102 L of distilled water for a dialysis patient. How many gallons of water is that?

Alinara [238K]

Answer:

63.36gallons

Explanation:

Given:

Volume of water used for dialysis = 2.4 x 10²L

Solution:

We are to convert from liters to gallons.

The conversion factor is shown below:

1L = 0.264gallons

To convert to litre:

since 1L = 0.264gallons

2.4 x 10²L = (2.4 x 10² x 0.264)gallons; 63.36gallons

3 0

3 years ago

How many moles are in a mass of 800.0 g of magnesium?

qaws [65]

Answer:

32.92 moles of Mg

Explanation:

To convert grams to moles (Or vice versa) of any chemical compound we need to use the molar mass of the substance (That is, how many grams weighs 1 mole of the chemical).

The magnesium, Mg, has a molar mass of 24.305g/mol. That means in 800.0g of Mg you have:

800.0g * (1mol / 24.305g) =

<h3>32.92 moles of Mg</h3>

7 0

2 years ago

A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v

mihalych1998 [28]

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

Number of moles = molarity x volume

Parameters given;

volume of solution = 500mL = 0.5L

molarity of solution = 0.5M

Number of moles = 0.5 x 0.5 = 0.25moles

Now to know the volume stock to take;

Volume of stock = $\frac{number of moles }{molarity}$

molarity of stock = 4M

volume = $\frac{0.25}{4}$ = 0.0625L or 62.5mL

4 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. The energy of one photon of this

Sedbober [7]

<u>Answer:</u> The energy of one photon of the given light is $3.79\times 10^{-19}J$

<u>Explanation:</u>

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

$\lambda$ = wavelength of light = $525nm=5.25\times 10^{-7}m$ (Conversion factor: $1m=10^9nm$ )

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{5.25\times 10^{-7}mm}\\\\E=3.79\times 10^{-19}J$

Hence, the energy of one photon of the given light is $3.79\times 10^{-19}J$

3 0

2 years ago