Express your answer using two significant figures. 1.7 km^2

Explain why some people use the term decimal system as a synonym for the metric system?

Luda [366]

Answer:

The term Decimal is coined from the word Decimus. Decimus is a Latin word which means ‘tenth’.

The Decimal system is a system of numbering that is based on multiples of tens. It is flexible and easy to convert from one metric to another.

The preference of the term over the Metric System is because the word Metric is related more to the measurement of lengths if we looked at the origin of the word which is French.

Cheers!

3 0

3 years ago

I will mark the brainiest answer

hoa [83]

The first one is Cesium chloride.. the second one is magnesium and sulfur so I’m not sure what’s going on with the answers

4 0

3 years ago

If there are 8g of carbon on the reactants side, how many g of carbon should you end with

Wewaii [24]

I think 8, because you have to end up with a balanced equation.

8 0

4 years ago

A student caries out an experiment and dissolves 10.0 g of sodium bromide. NaBr, in 200g of water. What is the molality of the s

Pavel [41]

Answer:

0.485mol/kg

Explanation:

Molality is defined as the ratio of the number of moles of a solute to the kilogram of a solvent which is usually water.

Mass of NaBr = 10g

Molar mass of NaBr= 23+80=103g/mol

Mole of NaBr = mass / molar mass=10g/103g/mol= 0.097mol

Molality= Mole of a solute/ mass of solvent in kg

Mole of a solute( NaBr)= 0.097mol; mass of solvent( water in kg)= 200g. To convert to kg: 200g/1000g= 0.2kg

Molality = 0.097mol/0.2kg = 0.485mol/kg

4 0

3 years ago

Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acet

Hoochie [10]

Answer:

0.182 moles of acetic acid are needed, this means 10.93 g.

0.318 moles of sodium acetate are needed, this means 26.08 g.

Explanation:

The Henderson–Hasselbalch (<em>H-H</em>) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + $log\frac{[A^{-} ]}{[HA]}$

In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.

From the problem we know that

[A⁻] + [HA] = 250 mM = 0.250 M eq. 1

We use the <em>H-H</em> equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + $log\frac{[A^{-} ]}{[HA]}$

$0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}]$ eq.2

Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:

1.74 [HA] + [HA] = 0.250 M

[HA] = 0.091 M

Then we calculate [A⁻]:

[A⁻] + 0.091 M = 0.250 M

[A⁻] = 0.159 M

Using the volume, we can calculate the moles of each substance:

moles of acetic acid = 0.091 M * 2 L = 0.182 moles
moles of sodium acetate = 0.159 M * 2 L = 0.318 moles

Using the molecular weight, we can calculate the grams of each substance:

grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
grams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 g

8 0

3 years ago