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3 years ago

What is reproduction

Chemistry

1 answer:

Zolol [24]3 years ago

3 0

The action or process of making a copy of something.

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A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

6 0

3 years ago

How is a gas's ability to fill a container different from that of a liquid or a solid?

svp [43]

Well a solid and a liquid is completely different from a gas. For a gas you need to have an air tight seal in order for the gas to stay in, or a special type of container so the gas doesn't decompose the container. And a liquid or a solid it's self explanatory. With some liquids you need special containers to hold the liquid so it doesn't decompose the container it's in.

8 0

3 years ago

A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti

Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)$

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

$n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2$

In such a way, the yielded moles of hydrobromic acid and chlorine are:

$n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2$

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0

3 years ago

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

Answer: The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

Explanation:

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

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

For nickel:

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

For nickel (II) oxide:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

For aluminium:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0

3 years ago

Are there any plans for controlling or removing zebra mussel

Sav [38]

Answer: no

Explanation:

4 0

3 years ago