Can someone help me PLEASE

what is the volume of a book that has a width of 10cm,a length that is 2 times the width,and a height that is half the width

Masteriza [31]

width=10cm

length=2xwidth=2x10=20cm

height=width/2=10/2=5

V=w*l*h=10*20*5=1000cm^3

5 0

3 years ago

Read 2 more answers

HURRY PLS!

marshall27 [118]

4 carbon chain is butane. With a triple bond it is butane. Since the triple bond is on the first carbon it can be called 1-butyne, or but-1-yne.

8 0

3 years ago

To what volume (in milliliters) should you dilute 100.0 mL of a 5.50 M solution of CaCl2 solution to obtain a 0.950 M solution o

Alexxx [7]

Dilution law is given as:

$M_{1}V_{1}=M_{2}V_{2}$ (1)

where, $M_{1}$ = molarity of initial concentrated solution.

$V_{1}$ = volume of initial concentrated solution.

$M_{2}$ = molarity of final diluted solution.

tex]V_{2}[/tex] = volume of final diluted solution.

Volume of initial concentrated solution of $CaCl_{2}$ = 100.0 mL

Molarity of initial concentrated solution of $CaCl_{2}$ = 5.50 M

Volume of final diluted solution of $CaCl_{2}$ = 0.950 M

Put the values in formula (1),

$5.50 M\times 100.0 mL=0.950 M\times V_{2}$

$V_{2} =\frac{5.50 M\times 100.0 mL}{0.950 M}$

$V_{2} =578.9 mL$

Hence, final diluted volume of the solution is $578.9 mL$ .

3 0

4 years ago

Assume that a milliliter of water contains 20 drops. How long, in hours, will it take you to count the number of drops

garri49 [273]

Answer:

126.18 hr

Explanation:

Data given:

1 mL of water = 20 drops

count rate = 10 drops/s

time in hours for one gallon = ?

Solution:

First we calculate number of mL (milliliter) of water in gallon

As we know

1 gallon = 3785.4 mL

As,

1 galon consist of 3785.4 mL of water, so now we count number of drops that contain 3785.4 mL of water

As we Know 1 mL water contain 20 drops then 3785.4 mL of water contain how many drops:

Apply unity formula

1 mL water ≅ 20 drops

3785.4 mL water ≅ X drops

Do cross multiplication

X drops of water = 20 drops x 3785.4 mL / 1 mL

X drops of water = 75708 drops

So, we come to know that one gallon contain 75708 drops of water and we have to calculate the time in hour to count these drops

First we calculate time in seconds

As we Know 10 drops water count in one second then how many seconds it will take to count 75708 drops

Apply unity formula

1 second ≅ 10 drops

X second ≅ 75708 drops

Do cross multiplication

X second = 1 second x 75708 drops / 10 drops

X second = 7570.8 second

So it take 7570.8 second to count 1 gallon water drops

Now convert seconds to hours

As,

60 seconds = 1 hr

7570.8 second = 7570.8 / 60 = 126.18 hr

So it take 126.18 hr to count 1 gallon water drops.

6 0

4 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,

$n(\text{CO}_2) = n(\text{C}_2\text{H}_6) \cdot (2 / 4) = 0.0510 \; \text{mol}$

$m(\text{CO}_2) = 0.0510 \; \text{mol} \times 44.01 \; \text{g} \cdot \text{mol}^{-1} = 2.24 \; \text{g}$

4 0

3 years ago