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1 year ago

1. Calculate the molar mass of sucrose (C12 H11 O22)?

Chemistry

1 answer:

ohaa [14]1 year ago

4 0

Answer:

=342g

Explanation:

atomic mass of C = 12g

atomic mass of H = 1g

atomic mass of O = 16g

Solution;

C12 H22 O11

= 12 (12) + 22 (1) + 11(16)

= 144+ 22 + 176

= 342g

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HCI + NaOH --> NaCl + H 20

Vikki [24]

Answer:

option D is correct

Explanation:

no of moles in 3 grams of HCL=3/36=0.08

if 1 mole of HCL require 1 mole of NaOH then 0.08 moles required 0.08 moles of NaOH

mass of 0.08 moles of NaOH=moles*molar mass=0.08*40=3.2 grams

so 3 grams are required in the reaction

6 0

2 years ago

Give the reason why the empty crucible should be heated before starting the experiment

sdas [7]

To make sure no cracks are in the crucible and also to remove any moisture present in the crucible by the process of heating.

3 0

2 years ago

How many moles of oxygen are required to react with methanol in order to produce 13 moles of formaldehyde?

ddd [48]

Answer:

6.5 moles of Oxygen are required

Explanation:

Based on the reaction:

CH3OH + 1/2 O2 → CH2O + H2O

1 mole of methanol reacts with 1/2 moles O2 to produce 1 mole of formaldehyde and 1 mole of water.

Thus, to produe 13 moles of formaldehyde, CH2O, are needed:

13 moles CH2O * (1/2mol O2 / 1mol CH2O) =

<h3>6.5 moles of Oxygen are required</h3>

3 0

2 years ago

What is the correct formula<br> for tin(II) chloride dihydrate?

seropon [69]

the answer is A. SnCl2 . 2H2O

3 0

2 years ago

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

erica [24]

<u>Answer:</u> The amount of $P_4O_{10}$ formed is 469.8 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

8 0

2 years ago