HELLLPPPP!!!!

How many hydrogen bonds are between cytosine and guanine

Dahasolnce [82]

Answer:

3

Explanation:

5 0

3 years ago

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Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

bija089 [108]

<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic \: mass }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 72 \times84.9 + 28 \times 87 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 6112.8 + 2436 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 8548.8 }{100} \\$

$\footnotesize \longrightarrow \: \bf Average \: atomic \: mass = 85.488 \: amu \\$

8 0

2 years ago

How can water and gravity work together to erode soil,sediment,and rock

Sunny_sXe [5.5K]

Ever seen a waterfall ? what do you think happens when all that water slams into the ground below ? same as when you turn a hosepipe on a flowerbed.

also, there's hydraulic cracking where water gets into cracks and freezes, splitting rocks apart.

fast moving water can carry sediment and roll rocks along the bottom. if the water slows down, it drops the rocks and heavier sediment. if it floods a field, say, then drains away, the grass will also filter smaller particles out, or the water might evaporate and leave fine sediment behind.

6 0

3 years ago

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If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac

STALIN [3.7K]

Answer: The theoretical yield of barium sulfate is 50.9 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sodium sulfate:

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol$

For barium chloride:

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

$\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol$

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

$Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl$

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.228=0.228mol$ of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = $\frac{1}{1}\times 0.228=0.228moles$ of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

$0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g$

Hence, the theoretical yield of barium sulfate is 50.9 grams

7 0

3 years ago

The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34 × 10-30 C-m. Ca

svetoff [14.1K]

The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 $\times$ 10^-5 V.

Explanation:

Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.

Given p = 1.47 D = 1.47 $\times$ 3.34 $\times$ 10^-30 = 4.90 $\times$ 10^-30.

V = 1 / (4π∈о) $\times$ (p cos(θ)) / (r^2)

where p is a permanent electric dipole,

∈ο is permittivity,

r is the radius from the axis of a dipole,

V is the electric potential.

V = 1 / (4 $\times$ 3.14 $\times$ 8.85 $\times$ 10^-12) $\times$ (4.90 $\times$ 10^-30 $\times$ 1) / (55.3 $\times$ 10^-9)^2

V = 1.44 $\times$ 10^-5 V.

8 0

3 years ago