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2 years ago

How many moles of KCl are in 28g of KCl?

Chemistry

2 answers:

eduard2 years ago

7 0

Answer:

0.38 moles KCl

Explanation:

(28 g KCl) / (74.55 g/mol KCl) = 0.38 moles KCl

netineya [11]2 years ago

6 0

Answer:

There are 0.38 moles of KCI in

28g of KCI.

Explanation:

This maybe correct.

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Which of the following combustion reactions is balanced correctly? A. C4H6 + 5.5O2 4CO2 + 3H2O B. C4H6 + 4O2 4CO2 + 3H2O C. C4H6

Airida [17]

You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.

<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O

element reactant side product side

C 4 4
H 6 3*2 = 6
O 5.5 * 2 = 11 4*2 + 3 = 11

Then, this equation is balanced.

</span>Do the same with the other equations if you want to verify that they are not balanced.

Answer: option A.

5 0

3 years ago

If eggs are so fragile, how do they actually protect the growing bird?

Step2247 [10]

Although birds' eggs appear to be fragile, they are in fact extremely robust. The oval shape applies the same rules of engineering as an arched bridge; the convex surface can withstand considerable pressure without breaking.

6 0

3 years ago

A 5.00-cm cube of magnesium has a mass of 217.501 g. What is the density of magnesiummetal?

Doss [256]

Answer:

d = 43.5 g/cm³

Explanation:

Given data:

Mass of magnesium cube = 217.501 g

Volume of magnesium cube = 5.00 cm³

Density of magnesium cube = ?

Solution:

Formula:

d = m/v

d = density

m = mass

v = volume

by putting values,

d = 217.501 g/ 5.00 cm³

d = 43.5 g/cm³

8 0

3 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

3 years ago

What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×

Ira Lisetskai [31]

$^{10}_5 B$
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

$5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}$

The answer is A. 1.67 × 10⁻²⁶ kg.

8 0

3 years ago