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3 years ago

The a-helices and b- sheets in protein make up its

Chemistry

1 answer:

Marta_Voda [28]3 years ago

5 0

Alpha helix and beta sheets are the secondary structure of protein

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Physical Science pre-Test 2020-2021

poizon [28]

Answer:

exothermic I think

Explanation:

IM TWKIGN THE TEST

4 0

2 years ago

Match the following items.

navik [9.2K]

Answer:

1. not affected by a magnet 1 liquid

2. mostly space 2 solid

3. flows freely but particles still attract 3 proton

4. positively charged particles 4 gas

5. free to move in all space 5 alpha particles

6. negatively charged particles 6 atoms

7. atoms nearly fixed in space 7 gamma ray 8. helium nucleus 8 electron (beta)

6 0

2 years ago

How many electrons are present in outermost shell of argon atom

sattari [20]

Answer:

there are 18 electrons in the outermost shell of argon atom

7 0

2 years ago

Read 2 more answers

What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen

Blizzard [7]

Answer: $Mg_{1}Cl_{2}O_{8}$

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg= $\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles$

Moles of Cl = $\frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles$

Moles of O = $\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = $\frac {0.45}{0.45}=1$

For Cl = $\frac {0.89}{0.45}=2$

For O= $\frac {3.58}{0.45}=8$

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is $Mg_{1}Cl_{2}O_{8}$

4 0

2 years ago

Please someone help me Balance this equation.

NeTakaya

Answer: This is an oxidation-reduction (redox) reaction:

3 C-II - 12 e- → 3 CII (oxidation)

4 CrVI + 12 e- → 4 CrIII (reduction)

C2H5OH is a reducing agent, K2Cr2O7 is an oxidizing agent.

7 0

2 years ago