Ask question

Ask question

All categories

4 years ago

11

i have a mixture of a tomato sauce, sugar, and carrots i bring a magnet close by. what does the magnet pick up and what is left

behind

1 answer:

MrRissso [65]4 years ago

3 0

Answer:

None.

Explanation:

Magnets only pick up metal, so in that case the magnet will not pick up any.

You might be interested in

What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?

dusya [7]

Answer:

$\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}$

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

$molarity = \frac {moles}{liters}$

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.

6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

$\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}$

Multiply by the given number of grams.

$78 \ g \ C_6H_12O_6 *\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}$

Flip the fraction and divide.

$78 \ g \ C_6H_12O_6 *\frac { 1 \ mol \ C_6H_12O_6}{180.156 \ g\ C_6H_12 O_6}$

$\frac { 78 \ mol \ C_6H_12O_6}{180.156 }= 0.432958102977 \ mol \ C_6H_12O_6$

2. Milliliters to Liters

There are 1000 milliliters in 1 liter.

$\frac {1 \ L }{ 1000 \ mL}$

Multiply by 2500 mL.

$2500 \ mL* \frac {1 \ L }{ 1000 \ mL}$

$2500 * \frac {1 \ L }{ 1000 }= 2.5 \ L$

3. Calculate Molarity

Finally, divide the moles by the liters.

$molarity = \frac {0.432958102977 \ mol \ C_6H_12O_6}{ 2.5 \ L}$

$molarity = 0.173183241191 \ mol \ C_6H_12O_6/L$

The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

$molarity \approx 0.17 \ mol \ C_6H_12O_6 /L$

1 mole per liter is also equal to 1 M.

$molarity = 0.17 \ M \ C_6H_12O_6$

3 0

3 years ago

What is the most common of all compounds are composed of molecules whose atoms are arranged in a straight chain, a branched chai

Gemiola [76]

<span>Sugar's carbon molecules can be arranged in a straight chain, a branched chain, or a ring.</span>

3 0

3 years ago

What is the molarity of a solution containing 8.0 moles of solute in 120mL of solution?

Nataliya [291]

Answer:
200/3 M which is approximately equal to 66.6667 M

Explanation:
Molarity is defined as the number of moles of solute per liter of solution.

It can be calculated as follows:
$molarity = \frac{number of moles of solute}{liters of solution}$

We are given that:
number of moles of solute = 8 moles
volume of solution = 120 ml = 0.12 liters

Substitute with the givens in the above equation to get the molarity as follows:
molarity = $\frac{8}{0.12} = \frac{200}{3} = 66.6667 M$

Hope this helps :)

3 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

(c) Δ $S_{sys}$ = 0 ; Δ $S_{sur}$ = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

6 0

3 years ago

Hsvsusvshssveuevsjsvsjdvsudvysvwkwhsjwvsgsbkwbsywvejwvsh

dybincka [34]

Sorry, I won't understand your words.

6 0

3 years ago

Read 2 more answers