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Rina8888 [55]
3 years ago
12

On a caterpillar's map, all distances are marked in millimeters. The caterpillar's map shows that the distance between two milkw

eed plants is 4, 012milim . What is this distance in kilometers ?
Chemistry
1 answer:
pychu [463]3 years ago
4 0

Answer:

0.004012km

Explanation:

Problem:

Conversion of millimeters(mm) to Kilometers:

value given = 4012mm

Here, we are converting from a submultiple unit to a multiple unit.

Millimeter depicts 10⁻³m and kilometer stands for 10³m

Now, we must find how many exponents will take us from 10⁻³ to  10³

careful examination shows that if we multiply a power of 10⁶ to 10⁻³ it will give a 10³:

   i.e 10⁻³ x 10⁶ = 10⁻³⁺⁶ = 10³

Therefore,

       10⁶mm = 1km

      4,012mm = \frac{4012}{1000000} = 0.004012km

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What is the mass of 2.50 moles of NaCl
garik1379 [7]

Answer:

The mass of 2,50 moles of NaCl is 146, 25 g.

Explanation:

First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:

Weight NaCl= Weight Na + Weight Cl=  23 g+ 35,5 g= 58, 5 g/ mol

1 mol ------ 58, 5 g

2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>

8 0
3 years ago
A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )
Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = \frac{0.1200}{1000}\times 50.0 moles = 0.00600 moles

Neutralization reaction (back titration): NaOH+HCl\rightarrow NaCl+H_{2}O

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = \frac{0.0980}{1000}\times 23.60 moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0
3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

6 0
3 years ago
CH3COOH(aq) + NaOH(aq) → H2O(l) + NaCH3COO(aq)
ser-zykov [4K]
I believe this is a double replacement. As you can see none of the chemical symbols are missing, there just "replaced". 
3 0
3 years ago
A cheetah can run 112km/h over a 100-m distance. what is this speed in meters per second
Naya [18.7K]
First convert the 112 km/hr ratio into m/s (meters per second). To do this you multiply 112 km with 1000 m/km (since there's 1000 m in one km). You get 112000 m. Then multiply 1 hr with 60 min/hr (since there's 60 min in one hr. You get 60 min, but you want seconds, so multiply 60 min with 60 s/min to get 3600 s. There you go! Your answer is the speed of 112000m/3600s, but you can simplify that to 31.11m/s (since the answer should be in ? meters per 1 second.

Also, the "100-m-distance" part of the question is just to throw you off, because one particular speed obviously stays constant over any distance. Hope that helps :)
7 0
3 years ago
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