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3 years ago

I need a person i can love and not get hurt

Chemistry

1 answer:

Katyanochek1 [597]3 years ago

7 0

Answer:

You'll find someone. There's a someone for everyone.

Explanation:

And I know how you feel.

You might be interested in

What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-

Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

Step 1: Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

Step 2: Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

Step 3: Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

Step 4: Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

4 0

3 years ago

A 10-ft moving truck has a volume capacity of 402 cubic feet, which is equivalent to 1.138 x 10⁷ milliliters. What mass of silve

steposvetlana [31]

Answer:

The mass of silver is $1.19\times 10^{14}\ Mg$

Explanation:

Given that,

The volume of the truck is 402 cubic feet or 1.138 x 10⁷ mm.

The density of silver is 10.49 g/mL.

We need to find the mass of silver that will fit in this moving truck. Density equals mass divided by volume. So,

$d=\dfrac{m}{V}\\\\\text{where m is mass}\\\\m=d\times V\\\\m=10.49\ g/mL\times 1.138 \times 10^7\ mL\\\\m=119376200\ g$

1 gram = 10⁻⁶ Megagram

$119376200=1.19\times 10^{14}\ Mg$

Hence, the mass of silver is $1.19\times 10^{14}\ Mg$

4 0

3 years ago

Before measuring the absorbance of a solution with the ocean optics spectrophotometer, it is important to run a blank sample of

Strike441 [17]

I think the most appropriate answer is: the solvent being used in the experiment
To correct for any light absorption not originating from the solute you will need to calibrate the tools with a solution that most similar to the sample.
Blank covete or standard solution can be used, but it was not ideal. By using the solvent as calibration, you can remove the reading from the solvent so your result only comes from the sample.

5 0

3 years ago

How many atoms are in 25.00 g of B.

iren2701 [21]

Answer:

$\boxed {\boxed {\sf 1.393 *10^{24} \ atoms \ B}}$

Explanation:

1. Convert Grams to Moles

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

Boron (B): 10.81 g/mol

Use this value as a ratio.

$\frac {10.81 \ g \ B }{1 \ mol \ B}$

Multiply by the given number of grams.

$25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}$

Flip the ratio so the grams of boron cancel out.

$25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}$

$25.00 *\frac {1 \ mol \ B }{10.81 }$

$\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B$

2. Convert Moles to Atoms

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

$\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

Multiply by the number of moles we calculated.

$2.312673451 \ mol \ B *\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

The moles of boron cancel.

$2.312673451 *\frac {6.02*10^{23} \ atoms \ B} {1 }$

$2.312673451 *6.02*10^{23} \ atoms \ B} =1.39269195*10^{24} \ atoms \ B$

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

$1.392\underline69195*10^{24} \ atoms \ B$

The 6 tells us to round the 2 to a 3.

$1.393 *10^{24} \ atoms \ B$

25.00 grams of boron is equal to 1.393*10²⁴ atoms.

6 0

3 years ago

What kind of bond would a metal and a nonmetal typically make?

lana66690 [7]

A) ionic bond.......

4 0

3 years ago

Read 2 more answers