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Hitman42 [59]
3 years ago
10

The organelle that is responsible for making glucose is the ____.

Chemistry
2 answers:
ivolga24 [154]3 years ago
7 0

The answer is mitochondria.

shtirl [24]3 years ago
5 0
Your answer would be mitochondria!
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The mass of cupric sulfate needed to make .5 liters of a .5M solution is
dalvyx [7]

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C

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Who want's to talk there is nothing to do​
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A liquid has an empirical formula CCl2, and a boiling point of 1 21 oC. When vapourised, the gaseous compound has a density of 4
Darya [45]

Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

  • molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

Pressure = 785 torr = 1.033 atm

density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

Learn more about molar mass of a gas at: brainly.com/question/26215522

6 0
2 years ago
Mercury has a mass density of 13.54 g/ml . how many milliliters would 100. grams occupy
frosja888 [35]
<span>7.39 ml For this problem, simply divide the mass of mercury you have by it's density. 100 g / 13.54 g/ml = 7.3855 ml Since we only have 3 significant digits in 100., you need to round the result to 3 significant digits. So 7.3855 ml = 7.39 ml</span>
6 0
3 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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