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3 years ago

6

The initial concentration of a solution of NaOH is 5.0 M. What volume of the concentrated stock solution is needed to make 200 m

L of .02 M solution? (YOU MUST INCLUDE UNITS IN YOUR ANSWER) *

1 answer:

sveticcg [70]3 years ago

4 0

Answer:

0.8 mL

Explanation:

Use the formula M1V1 = M2V2.

Plug the values in.

5(V1) = 0.02(200)

5V1 = 4

V1 = 0.8 mL

Hope this is right, correct me if I'm wrong!

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If you lived in Flagstaff, Arizona, how much salt (NaCl) would you have to add to your spaghetti water to get it to boil at 100

nasty-shy [4]

Answer:

Explanation:

This question is both theoretical and practical. While the theoretical aspect will be detailed fully here, the practical aspect will be provided as a form of guidance.

Water generally boils at 100°C when altitude (in feet) is 0. One of the colligative properties that occurs <u>when salt is added to water is that there is a boiling point elevation</u>(meaning an increase in boiling point). For instance, if 20g of salt is added to about 5.3 quarts of water, the boiling point of water will increase from 100°C to 100.04°C.

However, when the altitude/elevation of a place is about 7000 ft (like in Flagstaff, Arizona), water will boil at 95.3°C. In order to get 2 quarts of water to boil at 100°C in Flagstaff;

20g causes an increase in boiling point by 0.04°C (100°C to 100.04°C) in 5.3 quarts of water

What gram will increase the boiling point by same 0.04°C in 2 quarts

20g ⇒ 5.3

X ⇒ 2

5.3 X ⇒ 40g

X = 40 ÷ 5.3

X = 7.55g

Hence, 7.55g will cause an increase in boiling point by 0.04°C (from 100°C to 100.04°C) in 2 quarts of water

What mass of salt will increase the boiling point by 4.7°C (95.3°C to 100°C)

7.55g ⇒ 0.04

X ⇒ 4.7

X × 0.04 ⇒ 7.55 × 4.7

0.04X ⇒ 35.5

X = 887.5g

Hence, in order for the spaghetti water to boil at 100°C, 887.5g of salt needs to be added.

For the practical part of the question, some Kitchen scales have an accuracy of .25kg (250g) and some have an accuracy of .2 kg (200g) and some have an accuracy of .5kg (500g). The one your kitchen has will determine the amount of salt that you can measure. For example, if your kitchen scale/balance has an accuracy of 250g/0.25kg, then you can only measure 750g of the 887.5g (as the rest is 137.5g, which is not up to 250g of the scale's accuracy) of the required salt measurement. However, if you have a digital balance that can measure up to 2kg/2000g in one decimal place, that's the perfect balance to measure this salt.

5 0

4 years ago

Which action is the best example of a direct observation

luda_lava [24]

Since you didn't have any extra information about the question I'll be presenting an example from my own textbooks that I've used.

An example of a direct observation is listening to a cricket chirp at night, and counting the number of chirps per minute.

Direct Observation is where the evaulator watches the subject in their usual habitat without disrupting or altering it.

3 0

3 years ago

Read 2 more answers

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb

damaskus [11]

ANSWER IS D.

EXPLANATION:

The electronic configuration of neutral Rubidium atom is 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,5s1.

when a monoatomic ion is to be formed the Rubidium atom loses one electron from 5s orbital so the electronic configuration for the ion is 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6.

3 0

3 years ago

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g).

svetoff [14.1K]

Answer: 2.1 g mass of ozone( $O_{3}$ ) is predicted to form from the reaction of 2.0 g $NO_{2}$ in a car's exhaust and excess oxygen

Given information : Mass of $NO_{2}$ = 2.0 g and $O_{2}$ is in excess.

We need to calculate the mass of ozone ( $O_{3}$ )

Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

$NO_{2} + O_{2}\rightarrow NO + O_{3}$

Step 1 : Convert grams of $NO_{2}$ to moles of $NO_{2}$ .

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $NO_{2}$ = 46.0 g/mol

$Moles = \frac{2.0g}{46.0\frac{g}{mol}}$

Moles of $NO_{2}$ = 0.043 mol

Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

A mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

Moles of $O_{3}$ = $(0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}$

Moles of $O_{3}$ = $(0.043)\times \frac{(1 mol O_{3})}{(1)}$

Moles of $O_{3}$ = 0.043 mol

Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

Grams = $(0.043)\times (\frac{48 g O_{3}}{1})$

Grams = 2.1 g $O_{3}$

Note : The above three steps can also be done using a single step setup.

Grams of $O_{3}$ = $(2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}$

Grams of $O_{3}$ = $(2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}$

Grams of $O_{3}$ = 2.1 grams

4 0

3 years ago

How much energy is required to raise the temperature of 3 kg of iron from 20°

Vinvika [58]

Answer:

Option C (6750 J) is the appropriate response.

Explanation:

The given values are:

Mass of iron,

m = 3 kg

Specific heat capacity,

$c_A$ = 0.450

Temperature,

T = $25-20$

= $5^{\circ}C$

Now,

⇒ $Q=mc_A T$

On substituting the values, we get

⇒ $=3\times 0.450\times 5$

⇒ $=6.75$

i.e,

⇒ $=6750 \ J$

8 0

3 years ago