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koban [17]
2 years ago
10

Which two types of cells are involved in fertilization?

Chemistry
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:

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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
2 years ago
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