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3 years ago

You need to make up 500.0 ml of .5750 m glucose (180.15g/mol). What is the mass of the glucose should you measure out

Chemistry

1 answer:

Marianna [84]3 years ago

3 0

Answer:

Calculate the volume (in mL) of the 1.356 M stock NaOH solution needed to prepare 250.0 mL ... Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar ... g of glucose in enough water to make 500.0 mL of solution. • Step 2: Transfer 18.6 mL of glucose

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Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase

olchik [2.2K]

Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm

D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

7 0

3 years ago

Which argument best explains the charge of an atomic nucleus?

xeze [42]

The answer is "<span>An atomic nucleus is positively charged because it is composed of protons". An atomic nucleus actually contains nucleons which are made up of both protons and neutrons. Since neutrons are neutral or have no charge, the charge of an atomic nucleus mainly relies on the positive charge of the protons.</span>

6 0

3 years ago

Read 2 more answers

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

8 0

4 years ago

Assume that the reaction of aqueous hydrobromic acid solution and potassium hydroxide base undergoes a complete neutralization r

Tamiku [17]

Answer:

$HBr + KOH → KBr + H _{2} O$

$1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\ = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\ = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\ = 0.684 \: g$