What effect does earths tilt have on the planet

Please I need help with questions 1-4 and can you check if I plot it correct in to the punnet square for questions 1-4 and I nee

fgiga [73]

Number 1 is incorrect, the genotypes are given to you. You need to use GG and gg. The outcome would be 100% Gg.

Number 2 is incorrect, the genotypes are given to you. You need to use Gg and Gg. The outcome would be 25% gg, 25% GG, and 50% Gg.

Number 3 is incorrect, the genotypes are given to you. You need to use TT and tt. The outcome would be 100% Tt.

Number 4 is incorrect, the genotypes are given to you. You need to use RR and rr. The outcome would be 100% Rr.

Please read the directions and use the genotypes they give you! The information is all there for you, you just need to put it in correctly. If you're still having trouble understanding how to do this, feel free to message me and I'd be happy to help you.

8 0

3 years ago

How many moles of gas are in a 1.0 liter canister if the temperature of the canister is 100 K and the pressure is 100 atmosphere

ladessa [460]

Answer:

n = 12.18 moles

Explanation:

Given that,

The volume of a canister, V = 1 L

The temperature of the canister, T = 100 K

Pressure, P = 100 atm

We need to find the number of moles of gas. Let there are n number of moles. We know that,

PV = nRT

Where

R is gas constant, R = 0.0821 L*atm/mol*K

$n=\dfrac{PV}{RT}\\\\n=\dfrac{100\times 1}{0.0821 \times 100}\\\\n=12.18\ moles$

Hence, there are 12.18 moles of gas.

4 0

3 years ago

Pick the correct statement for the following isotope: a. 42Ca 42 is the mass number and 20 is the atomic number. b. 42 is the nu

kramer

Answer:

A

Explanation:

To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.

We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.

Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.

4 0

3 years ago

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0

3 years ago

If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,

rewona [7]

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

a)

If m1, V1 and D1 = mass, volume and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

Ans: Ball A will have a greater density than B

b)

If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

c)

If m1, V1 and D1 = mass, volume and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

d)

If m1, V1 and D1 = mass, volume and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.

5 0

3 years ago

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