How old is a piece of wood that has 0.7813% of the carbon-14 of a living organism

Polycarbonate plastic has a density of 1.2 g/cm3 . A photo frame is constructed from two 3.0 mm sheets of polycarbonate. Each sh

valina [46]

<span>440 g First, determine the volume of each sheet. And it's easier if each dimension is using the same unit of measure. So each sheet is 28 cm by 22 cm by 0.30 cm. Multiply them together 28 cm * 22 cm * 0.30 cm= 184.8 cm^3 Since we have 2 identical sheets, double the total volume 184.8 cm^3 * 2 = 369.6 cm^3 Now multiple the volume by the density 369.6 cm^3 * 1.2 g/cm^3 = 443.52 g Round the results to 2 significant digits since all of the given figures are only 2 significant digits long. 443.52 g = 440 g</span>

6 0

2 years ago

A fox spots a rabbit in a field. The fox begins to chase the rabbit, and the rabbit runs away. Which statement best describes th

Ira Lisetskai [31]

Answer:

2 I Think

Explanation:

8 0

2 years ago

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A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0

2 years ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi

jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be PO43- < NO3- < Na+

4 0

2 years ago

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