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2 years ago

Please help meee??????

Chemistry

1 answer:

balu736 [363]2 years ago

6 0

Answer:

Prophase

Metaphase

Anaphase

Telophase

Explanation:

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Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one spe

Arturiano [62]

A. Fe2O3 + 3CO= 2Fe+3CO2
Here element oxidised is CO or Carbon Monoxide, since oxygen is added.

B. 2HCl+2KMnO4+3H2C2O4=6CO2+2MnO2+2KCl+4H2O
Here Element reduced is 3H2C2O4, since Hydrogen is being added. Also KMnO4 is reduced, since Oxygen is removed.

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3 years ago

Why must we do the a lot of quantity urine

serg [7]

Answer:

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ExplanatioN:

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2 years ago

How do felsic igneous rocks

Svetradugi [14.3K]

Answer:

A. Felsic igneous rocks are less dense than mafic igneous rock

Explanation:

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2 years ago

1. 2.0 mL/s x 4.0 hrs = ? L<br><br> How do you do this conversion problem

Phantasy [73]

Answer:

29 L.

Explanation:

Hello!

In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:

$2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L$

Which turns out 29 L with 2 significant figures.

Best regards!

6 0

2 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago