Which of these claims about the way energy is transferred in Earth’s interior is supported by evidence ?

7. You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. But f

Vilka [71]

Answer:

58.61 grams

Explanation:

Taking The molecular weight of NaCl = 58.44 grams/mole

Determine how many grams of NaCl to prepare the bath solution

first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution

= ( 6 * 170 ) / 1000

= 1020 / 1000 = 1.020 moles

= moles of NaCl * molar mass of NaCl

= 1.020 * 58.44

= 58.61 grams

4 0

2 years ago

Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

Let's assume we have 1 mol of cacodyl, in that case we'd have 209.96 g of cacodyl and the following masses of its components:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

12.09 g H ÷ 1 g/mol = 12.09 mol H

149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

3 0

2 years ago

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

2 years ago

Mgcl2 has a concentration of 0.054 M in the ocean. How many grams are present in 25ml of sea water

d1i1m1o1n [39]

Answer:

0.129g MgCl2

Explanation:

For this we need to understand the concept of molarity.

Molarity is number of moles of solute/litres of solution

M=n/L

Here we are given molarity of 0.054M and volume of 25ml. we just plug this in formula to find moles of MgCl2

0.054=x/(25/1000) (we divided 25 by 1000 to convert it to litres of solution)

x=0.00135 moles of MgCl2 (we are not done yet the question asks for grams so to convert to grams we multiply by molar mass of MgCl2.)

0.00135molesMgCl2 x 95.211g MgCl2/1molMgCl2

= 0.129g MgCl2

3 0

2 years ago

5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is

GenaCL600 [577]

1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V = 7.9 atm

3 0

2 years ago

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