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2 years ago

10

Which two types of cells are involved in fertilization?

1 answer:

astra-53 [7]2 years ago

4 0

Which two types of cells are involved in fertilization?

$\sf\purple{D.\: egg\: cell;\: sperm \:cell}$ ✅

Gametes are an organism's reproductive cells. The female gametes are called ova or egg cells, and the male gametes are called sperm.
During fertilisation, the gametes (an egg and sperm) fuse to form a zygote.

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

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Which reagent is the limiting reactant when 0.700 mol al(oh)3 and 0.700 mol h2so4 are allowed to react?

lisov135 [29]

Hffgbhv vice uvula Gn kick kill is Lal Lal lam Lal lisp loco num Mann ight isms genus in I icon 25

3 0

3 years ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.

6 0

3 years ago

Qué valores rescatas de la labor que realizan los bomberos

sertanlavr [38]

Explanation:

Honor:Cumplimos con nuestros deberes con entusiasmo y entrega. Honradez:Respeto a los demás, coherencia física e intelectual en lo que pertenece a cada persona y convicción de defenderlo. Humanismo:Preocupación por lo social y humano. ... Respeto:Tratar humanamente a las personas.

8 0

2 years ago

B) calculate the ratio of moles of h2o to moles of anhydrous kal(so4)2. note: report the ratio to the closest whole number. (sho

Sindrei [870]

Answer: -

Mass of Hydrated KAl(SO₄)₂ = 2.0 g

Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol

Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A

The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating

= H g

Moles of water released = $\frac{H g}{18 g/mol}$

Moles of anhydrous KAl(SO₄)₂ = $\frac{A g}{258.20 g/mol}$

Required ratio = $\frac{moles of water}{moles of anhydrous KAl(SO4)2}$

5 0

3 years ago

a 100 gram sample of iron ore was found to contain 44g of iron. How many grams of iron are in a 450 gram sample of the ore?

Sphinxa [80]

Answer:

There are 198 grams of iron in a 450-gram sample of the ore.

Explanation:

From the given information:

Given that:

100-gram sample of iron ore is found in 44g of iron.

Thus. the grams of iron that can be found in a 450-gram sample of the iron ore can be computed as follows:

100g of iron ore = 44g of iron

450 gram of iron ore = x g of iron

x(g) of iron = ( 44g of iron × 450 gram of iron ore) / 100g of iron ore

x(g) of iron = 198 grams of iron

Thus, there are 198 grams of iron in a 450-gram sample of the ore.

7 0

3 years ago