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3 years ago

If a cell is 80% water and the outside environment is 90% water. What is likely to happen?

Chemistry

1 answer:

Svet_ta [14]3 years ago

6 0

Water will rush into the cell, because water likes everything to be equal. This process is called diffusion (osmosis).

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How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?

natali 33 [55]

I think the answer is D

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3 years ago

Which statement best describes saturation?

sp2606 [1]

statement c would be correct I think.

8 0

3 years ago

Read 2 more answers

J.J. Thomson's model of the atom includes all BUT ONE of these features. That is

kakasveta [241]

Answer:

Explanation:

usa testprep!!!!!

8 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures.

VladimirAG [237]

Given:

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

To determine:

Theoretical yield of calcium phosphate, Ca3(PO4)2

Explanation:

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 = 96.1 g

Molar mass of Ca(NO3)2 = 164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g

7 0

3 years ago

Read 2 more answers

One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w

4vir4ik [10]

Answer:

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0

4 years ago