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3 years ago

1. What is the correct order of the water cycle?

Chemistry

1 answer:

irga5000 [103]3 years ago

3 0

C.
Evaporation
Condensation
Precipitation

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Histones are proteins that control gene function by attaching through salt links to exterior regions of DNA. Name an amino acid

sweet-ann [11.9K]

Histones are proteins that control gene function by attaching salt links to exterior regions of DNA. Argentines is an amino acid whose side chain is often found on the exterior of histones

A chromosome's structural support is provided by a protein called a histone. Long DNA molecules found on each chromosome must fit into the cell nucleus. This is accomplished by the DNA wrapping around histone protein complexes, giving the chromosome a more compact form.

The eight-protein complex known as a histone octamer is what makes up the nucleosome core particle. Each of the four main histone proteins is present in two copies in this structure (H2A, H2B, H3 and H4). When a tetramer, which has two copies of both H3 and H4, interacts with two H2A/H2B dimers, the octamer forms.

To learn more about histone please visit-
brainly.com/question/13036208
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7 0

2 years ago

What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?

Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g. 1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

6 0

3 years ago

Read 2 more answers

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

Neporo4naja [7]

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Consider the following serial dilutions.

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is diluted to 250 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM :

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM :

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is diluted to 500 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

7 0

3 years ago

4 moles of monoatomic ideal gas is compressed adiabatically causing the temperature to increase from 300 K to 400 K. Calculate t

yawa3891 [41]

Answer:

the work done on the gas is 4,988.7 J.

Explanation:

Given;

number of moles of the monoatomic gas, n = 4 moles

initial temperature of the gas, T₁ = 300 K

final temperature of the gas, T₂ = 400 K

The work done on the gas is calculated as;

$W = \Delta U = nC_v(T_2 -T_1)$

For monoatomic ideal gas: $C_v = \frac{3}{2} R$

$W = \frac{3}{2} R \times n(T_2-T_1)$

Where;

R is ideal gas constant = 8.3145 J/K.mol

$W = \frac{3}{2} R \times n(T_2-T_1) \\\\W = \frac{3}{2} (8.3145) \times 4(400-300) \\\\W = \frac{3}{2} (8.3145) \times 4(100)\\\\W = 4,988.7 \ J$

Therefore, the work done on the gas is 4,988.7 J.

4 0

3 years ago

Ask What do you think will happen when different liquids are mixed with water?

Elza [17]

Answer:

could create chemical reaction

Explanation:

5 0

3 years ago