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mars1129 [50]
3 years ago
7

What does a cell do to a substance in Endocytosis

Chemistry
1 answer:
8090 [49]3 years ago
3 0
Endocytosis is a type of bulk transport which involves the movement of large particles through the membrane in and out
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If your end product is 1.5 moles of KMnO 4 how many moles of manganese oxide were used in the reaction? The equation for the pro
vovikov84 [41]

Answer:

1.5 moles

Explanation:

The equation of the reaction is given as:

2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2

From the equation,

2 moles of MnO2 produces 2 moles of KMnO4

x moles of MnO2 would produce 1.5 moles of KMnO4

2 = 2

x = 1.5

Solving for x;

x = 1.5 * 2 / 2

x = 1.5 moles

4 0
3 years ago
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
Vilka [71]

Answer:

147.2g

Explanation:

The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density

8 0
3 years ago
A 500 gram piece of metal had a volume of 2.75cm3. What is it’s density
beks73 [17]

Answer:

181.82 g/cm3

Explanation:

density is mass / volume so it is 500 / 2.75=181.82 g/cm3

8 0
3 years ago
Which of the following correctly describes the role of scientific models in the advancement of technology? Models represent the
Nikitich [7]
Models help us better understand the phenomena
3 0
3 years ago
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
3 years ago
Read 2 more answers
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