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2 years ago

HELP ASAP, PLEASE!

1 answer:

algol132 years ago

7 0

A) James Cook.
B) He put his sailors on a strict diet to see if they would get scurvy.
C) Sauerkraut.
D) He told others of this diet and that none of his sailors died of scurvy.
E) Chemicals can be found almost anywhere and almost anyone can be a scientist.

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How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol).

Scilla [17]

Answer:

Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O → 3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

HCl : BCl₃

3 : 1

0.27 : 1/3×0.27 = 0.089 mol

6 0

2 years ago

A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model

san4es73 [151]

c is the correct answer

4 0

2 years ago

If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate

Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:

$\frac{Mass of glue}{Mass of glue + Mass of water}$

$\frac{10 g}{10 g + 13 g }$

$\frac{10 g}{23 g }$

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

$\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}$

$\frac{ 8 g}{ 8 g + 11 g }$

$\frac{8 g}{19 g }$

3 0

2 years ago

A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak

baherus [9]

Answer:

The volume of the sample is 17.4L

Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>

8 0

1 year ago

What is the reaction energy Q of this reaction? Use c2=931.5MeV/u. Express your answer in millions of electron volts to three si

emmasim [6.3K]

Answer:

Energy= 2.7758 × 10^-11 J ;

71.112×10^-6 kJ.

Mass defect in Kilogram= 3.0885×10^-28 kg.

That is; 3.1×10^-28 kg(to two significant figure).

Explanation:

(Note: Check equation of reaction in the attached file/picture).

STEP ONE: we have to calculate the Mass defect.

Mass defect= Mass of reactants -- Mass of products.

Mass of the products: (140.9144+91.9262+3.060) u.

= 235.8666 u.

Mass of reactants: (1.0087+235.0439) u= 236.0526 u.

Therefore, the Mass defect= (236.0526 -- 235.8666) u

= 0.1860 u.

STEP TWO: Converting the Mass defect to energy;

0.18860 × 1.6605 × 10^27 kg

= 3.0885× 10^-28 kg

STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.

E=Mc^2.

Where E= energy released, c= speed of light, M= Mass.

Slotting in the values;

E= 3.0885×10^-28 kg × 9×10^16 m/s.

E=2.7758 × 10^-11 J.

Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.

=7.1112×10^-10 J

= 71.112×10^6 kJ.

3 0

2 years ago