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Rus_ich [418]
3 years ago
7

3. In all cells, genetic materials are stored in the

Chemistry
2 answers:
Rainbow [258]3 years ago
5 0
Nucleus c
Because it is the powerhouse of the cell
jasenka [17]3 years ago
3 0
The answer is C the nucleus
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PLS HELP Students in a chemistry class added 5 g of Zinc (Zn) to 50 g of hydrochloric acid (HCl). A chemical reaction occurred t
Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be  Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is  5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

is 0.154 g.  Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =

 10.42 g

4 0
2 years ago
What type of energy changes take place when a candle burns
frutty [35]
The two types of energy changes that occur are heat and light changes.
7 0
3 years ago
Read 2 more answers
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
What is the molarity of a 2.0L solution that was made up with 4.0 moles of NaCl?
Finger [1]

Answer:

[NaCl]: 2M

Explanation:

This solution is made of NaCl therefore:

Our solute is NaCl

Moles of solute: 4

Our solution's volume is 2L

Molarity are the moles of solute contained in 1L of solution (mol/L)

[NaCl]: 4 mol /2L = 2M

We can also make a rule of three:

In 2 L we have 4 moles of solute

So, In 1 L we must have (1 . 4) /2 = 2 M

8 0
3 years ago
A student uses a calorimeter to determine the enthalpy of dissolving for ammonium nitrate. The student fills a calorimeter with
ale4655 [162]
Enthalpy change during the dissolution process = m c ΔT,

here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
                                                                                 = -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)

Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>

7 0
3 years ago
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