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Harrizon [31]
3 years ago
7

Modeling Mitosis and Meiosis

Chemistry
1 answer:
Helen [10]3 years ago
7 0

I've tried this like, probably 10 times, it never worksbut ig we'll see

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A copper coin with a mass of 6.30 grams changes temperature in the sunshine from 20.0°C to 32.0°C. How much energy is transferre
Mariana [72]

The energy transferred is  28.5  j


<em><u>calculation</u></em>

energy is calculate  MCΔT  formula  where,

M(mass)= 6.30  grams

C(specific heat capacity)= 0.377 j/g

ΔT(change in temperature)= 32.0c- 20 c= 12 c


Energy is therefore= 6.30 g   x 0.377  j/g /c x   12 c =<u>28.5 j</u>

8 0
4 years ago
Read 2 more answers
2 AgNO3(aq) + CaCl2(aq) -----&gt; 2 AgCl(s) + Ca(NO3)2(aq)
yan [13]

5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.

Explanation:

The balanced equation:

2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)

data given:

volume of AgNO3 = 0.2 L

molarity of AgNO3 = 0.200 M

atomic weight of AgCl= 143.32 gram/mole

from the formula, number of moles can be calculated

Molarity = \frac{number of moles}{volume in litres}

number of moles of AgNO3 = 0.04

From the reaction:

2 moles of AgNO3 reacts to form 2 moles of AgCl

0.04 moles of AgNO3 reacts to form x mole of AgCl

\frac{2}{2} = \frac{x}{0.04}

= 0.04 moles of AgCl is formed

mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl

mass of AgCl = 0.04 x 143.32

                       = 5.732 grams of AgCl is formed.

4 0
3 years ago
Metal M has only one oxidation state and forms a compound with the formula MHCO3. Which formula is correct? A.MNO3 b.MNH4 c.M(NO
never [62]

Answer:

The compound with the correct formula is;

A. MNO₃

Explanation:

The number of oxidation states in the metal, M = One oxidation state

The formula of the compound formed by the metal, M = MHCO₃

We note that the ion HCO₃⁻, known as hydrogen carbonate has an oxidation number of -1

Similarly nitrate, NO₃⁻ has an oxidation number of -1, therefore, the metal M can form similar compound formed with HCO₃⁻ with nitrate, and we have;

The possible compounds formed by the metal 'M' includes MHCO₃ and MNO₃.

5 0
3 years ago
The reaction system: POCl3(g) POCl(g) + Cl2(g) is at equilibrium. Which of the following statements describes the behavior of th
mars1129 [50]

Answer: (B) The reverse reaction will proceed to establish equilibrium.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

POCl_3(g)\leftrightharpoons POCl(g)+Cl_2(g)

Given : when POCl is added to the system

If the concentration of POCl that is the product is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of POCl takes place. Therefore, the equilibrium will shift in the left or backward direction.

Thus the reaction will shift towards reactants to reestablish equilibrium and thus partial pressure of POCl and Cl_2(g)  decreases and the partial pressure of POCl_3 increases.

8 0
3 years ago
Consider the reaction in the lead-acid cell pb(s) pbo2(s) 2 h2so4(aq) 2 pbso4(aq) 2 h2o(l) for which e°cell = 2. 04 v at 298 k.
aleksley [76]

δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

The lead (Pb) in the lead storage battery act as an anode

<h3>What is Voltaic Cell?</h3>

A voltaic cell or galvanic cell is an electrochemical cell that uses chemical reactions to produce electrical energy.

<h3>Parts of voltaic cell</h3>

Anode where oxidation occurs Cathode where reduction occurs.

Therefore, Pb acts as anode and PbO2 act as a cathode.

Given,

E° cell = 2.04v

Faraday constant= 96485 C/mol

Moles of electron transferred n= 2

∆G = -nFE°

= -2 × 96485 × 2.04

= 393.658kJ

Thus we concluded that the δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

learn more about voltaic cell :

brainly.com/question/1370699

#SPJ4

8 0
2 years ago
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