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Stells [14]
3 years ago
14

In roses, the color red is dominant (R), and the color white is recessive (r). Based on the Punnett Square, what is the probabil

ity ratio that the roses will be white?
Chemistry
1 answer:
anastassius [24]3 years ago
4 0

Answer:

0.25

Explanation:

From the given information:

The dominant trait R = red color; &

The recessive trait r = white

If a heterozygous traits Rr  crosses with a dominant trait RR;

The Punnet square will be as follows:

              R               r

R            RR             Rr

R            RR             Rr

Thus all the flowers are red and there is 50% dominant red rose flowers (RR) and 50% heterozygous red rose flowers (Rr).

Let assume that two heterozygous traits (Rr) cross together; Then the punnet square is:

          R         r

R        RR        Rr

r          Rr        rr

From the above punet square, we have:

0.25 dominant red rose flowers

0.5 heterozygous red rose flowers;

0.25 recessive white rose flowers.

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When the pressure that a gas exerts
Ronch [10]

Answer:

The pressure changes from 2.13 atm to 1.80 atm.

Explanation:

Given data:

Initial pressure = ?

Final pressure = 1.80 atm

Initial temperature = 86.0°C (86.0 + 273 = 359 K)

Final temperature = 30.0°C (30+273 =303 K)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

P₁ = P₂T₁ /T₂

P₁ = 1.80 atm × 359 K / 303 K

P₁ = 646.2 atm. K /303 K

P₁ = 2.13 atm

The pressure changes from 2.13 atm to 1.80 atm.

5 0
3 years ago
Read 2 more answers
Attempt 1
Vsevolod [243]

Answer:

12.01

Explanation:

(12.00*98.93% + 13*1.07%) /100% = 12.01

3 0
3 years ago
What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

6 0
3 years ago
Que materias o asignaturas disciplinarias consideras que están
zalisa [80]

Las áreas temáticas que pueden estar involucradas en el cuidado del agua son;

Biología

Química

Geología

oceanografía

Geografía

Conservar y compartir es una parte integral de la ciencia ambiental. La ciencia ambiental en sí misma es multidisciplinaria.

Varias áreas temáticas deben estar involucradas en la conservación del medio ambiente.

Estas áreas temáticas incluyen;

Biología

Química

Geología

oceanografía

Geografía

Aprende más: brainly.com/question/10929764

5 0
2 years ago
Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
dmitriy555 [2]

<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

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

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

  • <u>For aluminium oxide:</u>

Given mass of aluminium oxide = 12.5 kg = 12500 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

  • <u>For NaOH:</u>

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

  • <u>For HF:</u>

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol

For the given chemical reaction:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with (6\times 122.6)=735.6mol of sodium hydroxide and (12\times 122.6)=1471.2mol of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

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

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = \frac{2}{1}\times 122.6=245.2mol of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

7 0
3 years ago
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