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3 years ago

Amniotic egg is an egg that contains a f fluid-filled nucleus and a yolk sac

Chemistry

2 answers:

crimeas [40]3 years ago

5 0

So for that one I’m going to go with True

rodikova [14]3 years ago

4 0

The answer for this is TRUE

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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ella [17]

Answer :

The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of $Na_2CO_3$ = 4.25 g

Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

6 0

4 years ago

How many grams are there in 21.3 moles of baco3?

Alex787 [66]

Mass of CaCO3= mol x molar mass
= 21.3 mol x 197.335 g/mol
4203.2355g
=4.2032355g

8 0

4 years ago

AGO H2SO4 (aq) Al(SO4)3(aq) H28) fe Cl218) FeCl38<br>

Rina8888 [55]

Answer:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

2Fe + 3Cl2 → 2FeCl3

Explanation:

1. (SO4) 3 you see this 3 it means that 3 must be behind H2SO4. So now it's 3H2SO4.

2. If 3 is now behind one H2, it must be behind the other.

So now it's 3H2.

3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.

1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.

2. Since it is now 2FeCl3, we know that there must be 2 in the second Fe. It's 2Fe now.

7 0

3 years ago

Chemical reaction examples

Dmitry_Shevchenko [17]

An example is when a science volcano erupts with vinager and baking soda which explode to make a chimechal reaction

8 0

4 years ago

2 H2 (g) + O2 (g) --> 2 H2O (l)

goblinko [34]

Answer:

2H2(g]+O2(g]→2H2O(l]]. Notice that the reaction requires 2 moles of hydrogen gas and 1

Explanation:

5 0

3 years ago