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3 years ago

Please help will mark brainliest!

Chemistry

1 answer:

Llana [10]3 years ago

8 0

Answer:

1. desert

2. grasslands

3. rainforest

4. tundra

5. temperate deciduous forest

6. coniferous forest

Explanation:

educated guess, good luck

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You wish to make a 0.405 M hydrochloric acid solution from a stock solution of 3.00 M hydrochloric acid. How much concentrated a

Vesnalui [34]

Answer:

Explanation:

We have to make 100 mL 0f .405M HCl from 3 M solution of HCl .

volume of 3M to be taken required . Let this volume be V litre .

V litre of 3M will contain 3 V moles of HCl .

moles contained by 100 mL of ,405 HCl = .405 x .1 = .0405 moles .

So 3V = .0405

V = .0405 / 3 litre

= .0405 x 1000 / 4 mL

= 10.125 mL

So we have to take 10.125 mL of 3M HCl and add water to them to make its volume 100 mL .

4 0

2 years ago

If the temperature outside is 75oF and the dew point is 75oF what is the humidity?

Serjik [45]

Answer:

in this file you will find it https://xlbrands.com/wp-content/uploads/2014/06/DEW-POINT-CALCULATION-CHARTenglish.pdf

Explanation:

8 0

2 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon d

Korolek [52]

Answer:The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

Explanation:

Let the molecular formula of the oxide of metal be $X_2O_y$

$X_2O_y+yCO\rightarrrow yCO_2+2X$

Mass of metal product = 1.68 g

Moles of metal X = $\frac{1.68 g}{55.9 g/mol}=0.03005 mol$

1 mol of metal oxide produces 2 moles of metal X.

Then 0.03005 moles of metal X will be produced by:

$\frac{1}{2}\times 0.03005 mol=0.01502 mol$ of metal oxide

Mass of 0.01502 mol of metal oxide = 2.40 g (given)

$0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g$

y = 2.999 ≈ 3

The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

8 0

3 years ago

Name four types of pathogens that could cause infectious disease??

Kryger [21]

Bacteria, virus, fungi, and protists are the four pathogens.

3 0

2 years ago