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3 years ago

What is a good title for this chart?

Chemistry

1 answer:

Elina [12.6K]3 years ago

8 0

Answer:

pH of the acid

Explanation:

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A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50

Elden [556K]

Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

Height of the of the section of bar = h

Area of Equilateral triangular = $\frac{\sqrt{3}}{4}a^2$

a = 2.50 inches

Cross sectional area of the steel mass = A

$A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2$

$V = 2.71 inches^2\times h$

Density of the steel = d = $7.70 g/cm^3$

$1cm^3 = 0.0610237 inches^3$

$d=\frac{m}{v}$

$\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}$

$h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}$

h = 2.92 inches

The section of the bar is 2.92 inches.

3 0

3 years ago

Anna71 [15]

Na 1s²2s²2p⁶3s¹
↓ - e⁻
Na⁺ 1s²2s²2p⁶ 2+2+6=10 e⁻

10 electrons are in sodium ion Na⁺

4 0

3 years ago

What is a number bond.

GenaCL600 [577]

Explanation:

number bonds are pairs of numbers that can be added together to make another number e.g. 4 + 6 = 10. They are some of the most basic and most importantparts ofmaths for children to learn

3 0

2 years ago

i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t

Ronch [10]

Assuming that the number of mols are constant for both conditions:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Now you plug in the given values. V_1 is the unknown.
$\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}
$
Separate V_1
$V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }$
V= 162.782608696 L

There are 2 sig figs

V= 160 L

3 0

3 years ago

Calculate ΔHrxn for the following reaction: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/

sesenic [268]

Answer:

The answer to your question is: ΔHrxm = -23.9 kJ

Explanation:

Data:

2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ (1)

CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ (2)

Reaction:

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

We invert (1) and change the sign of ΔH

Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ

We multiply (2) by 3

3( CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ) (2)

3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ

We add (1) and (2)

Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ

3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ

Fe2O3(s) + 3CO(g)+3/2O2(g) → 2Fe(s)+3/2O2 + 3CO2(g)

Simplify

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) and ΔHrxm = -23.9 kJ

6 0

3 years ago