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3 years ago

Pls help

1 answer:

3 0

The Cambrian period, part of the Paleozoic era, produced the most intense burst of evolution ever known. The Cambrian Explosion saw an incredible diversity of life emerge, including many major animal groups alive today. Among them were the chordates, to which vertebrates (animals with backbones) such as humans belong. These included brachiopods, which lived in shells resembling those of clams or cockles, and animals with jointed, external skeletons known as arthropods—the ancestors of insects, spiders, and crustaceans. These toughened-up creatures represented a crucial innovation: hard bodies offering animals both a defense against enemies and a framework for supporting bigger body sizes.

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What two factors must be held constant for density to be a constant ratio?

Neko [114]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

Kindly mark it branliest if the answer is little bit satisfying.

7 0

3 years ago

How much charged is required to reduced 2 moles of cr6+ to cr3+

BartSMP [9]

Answer:

6e⁻

Explanation:

Charge in reducing 2 mole of Cr⁶⁺ to Cr³⁺

Cr⁶⁺ + xe⁻ → Cr³⁺

The numbers 6+ and 3+ on the chromium atoms are the oxidation number of the atom.

It shows the number of electrons that has been lost by chromium in an oxidation state.

Cr⁶⁺ shows that it has lost 6 electrons. When an atom loses electrons, the number of protons becomes more. This makes it positively charged in nature.

This analogy goes for Cr³⁺

Therefore, in going from reactants to products, chromium

6+ → 3+

In this kind of expression, the number of atoms must be conserved and the charges too;

2Cr⁶⁺ + xe⁻ → 2Cr³⁺

given 2 moles of Cr;

To balance the charge;

2(6+) + x(-) = 2(3+)

12 - x = 6

x = 6

A charge of 6e⁻ is required to reduce 2 mole of Cr⁶⁺ to Cr³⁺

6 0

4 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$

Mass of 44 moles of water ;

$=44mol\times 18g/mol=792g$

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

3 0

3 years ago

You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$

Best regards.

4 0

4 years ago

H20 + C02...................

icang [17]

Answer:

NICE

Explanation:

TOES

7 0

3 years ago