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2 years ago

Which of the following cooking materials does NOT conduct heat?

Chemistry

2 answers:

SVETLANKA909090 [29]2 years ago

5 0

The answer to your question is D because wood does not conduct heat

kherson [118]2 years ago

3 0

Can be produced from a variety of material, including , it’s at a C or D.

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space capsules operate with an oxygen content of about 34%. assuming a total pressure of 780 mm Hg in the space capsule, what is

Lemur [1.5K]

265.2 mmHg is the partial pressure of oxygen in 780 mmHg of total pressure.

Explanation:

The partial pressure of a gas is defined as the individual pressure of the gas in total mixture. In an ideal gas all the constituent gases have partial pressure some of which will give total pressure of the gas.

The partial pressure of a gas is calculated by

total pressure x mole fraction of the gas.

Mole fraction of the oxygen present is 0.34 as it is 34% of the total gas.

$\frac{34}{100}$ = 0.34 is the mole fraction

Total pressure is given as 780 mm Hg

The partial pressure can be calculated using the above formula:

Putting the values in equation:

780 x 0.34

= 265.2 mm Hg is the partial pressure of oxygen.

7 0

2 years ago

Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants

svet-max [94.6K]

Answer:

154.0831 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Mg$

Given mass = 41.0 g

Molar mass of $Mg$ = 24.31 g/mol

Moles of $Mg$ = 41.0 g / 24.31 g/mol = 1.6865 moles

Given: For $FeCl_3$

Given mass = 175 g

Molar mass of $FeCl_3$ = 162.2 g/mol

Moles of $FeCl_3$ = 175 g / 162.2 g/mol = 1.0789 moles

According to the given reaction:

$3Mg_{(s)}+2FeCl_3_{(s)}\rightarrow 3MgCl_2_{(s)}+2Fe_{(s)}$

3 moles of Mg react with 2 moles of $FeCl_3$

1 mole of Mg react with 2/3 moles of $FeCl_3$

1.6865 mole of Mg react with (2/3)*1.6865 moles of $FeCl_3$

Moles of $FeCl_3$ = 1.1243 moles

Available moles of $FeCl_3$ = 1.0789 moles

Limiting reagent is the one which is present in small amount. Thus, $FeCl_3$ is limiting reagent. (1.0789 < 1.1243)

The formation of the product is governed by the limiting reagent. So,

2 moles of $FeCl_3$ gives 3 moles of magnesium chloride

1 mole of $FeCl_3$ gives 3/2 moles of magnesium chloride

1.0789 mole of $FeCl_3$ gives (3/2)*1.0789 moles of magnesium chloride

Moles of magnesium chloride = 1.61835 moles

Molar mass of magnesium chloride = 95.21 g/mol

Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g

6 0

2 years ago

How many ml of a 14.0 m nh3 stock solution are needed to prepare 200 ml of a 4.20 m dilute nh3 solution? hints how many ml of a

Fudgin [204]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

14 M x V1 = 4.20 M x 200 mL

V1 = 60 mL needed of the concentrated solution

5 0

2 years ago

Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp

sergiy2304 [10]

Explanation:

The given data is as follows.

$t_{m}$ = 30.0 sec, $t_{r1}$ = 5 min = $5 \times 60 sec$ = 300 sec

$t_{r2}$ = 12.0 min = $12 \times 60 sec$ = 720 sec

Formula for adjusted retention time is as follows.

$t'_{r} = t_{r} - t_{m}$

= 300 sec - 30.0 sec

= 270 sec

$t'_{r2}$ = 720 sec - 30 sec

= 690 sec

Formula for relative retention ( $\alpha$ ) is as follows.

$\alpha = \frac{t'_{r2}}{t'_{r1}}$

= $\frac{690 sec}{270 sec}$

= 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0

2 years ago

Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon m

weeeeeb [17]

On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ 0,082 L·atm/K·mol · 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.

6 0

2 years ago

Read 2 more answers