What is the answer to 1 and 2?

Which of the following are the types of RNA?

Andrew [12]

Answer:

Explanation:

Of the numerous sorts of RNA, the three most well-known and most commonly examined are delivery person RNA (mRNA), exchange RNA (tRNA), and ribosomal RNA (rRNA), which are show in all living beings. These and other sorts of RNAs essentially carry out biochemical responses, comparative to proteins.

5 0

3 years ago

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In CF4 and NF3, the ___electron groups on the central C and N atoms have a ___ arrangement. The shapes of the molecules are dete

enyata [817]

Answer:

Explanation:

In CF4 and NF3, the valence electron groups on the central C and N atoms have a tetrahedral arrangement. The shapes of the molecules are determined by the number of bonding and nonbonding of electrons: since CF4 has four bonded atom(s) and zero lone pair(s) of electrons, the shape is tetrahedral.

7 0

3 years ago

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Anton van Leeuwenhoek used a microscope to observe

Goshia [24]

B. He was the first person to observe and identify living cells.

4 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures. Calcium nitrate reacts with sodium phospha

RoseWind [281]

6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).

<h3>How we calculate mass from moles?</h3>

Mass of any substance can be calculated by using moles as:

n = W/M, where

W = required mass

M= molar mass

Given chemical reaction is:

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

From the stoichiometry it is clear that:

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

Given mass of Ca(NO₃)₂ = 96.1g

Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles

So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂

Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:

W = 0.0196mole × 310g/mole = 6.07 grams

Hence, 6.07 grams is the theoretical yield of calcium phosphate.

To know more about moles, visit the below link:

brainly.com/question/15373263

7 0

2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

3 years ago

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