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2 years ago

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What are common names for antimony ? for science

1 answer:

dusya [7]2 years ago

4 0

Antimony is a chemical element with symbol Sb (from Latin: stibium) and atomic number 51. A lustrous gray metalloid, it is found in nature mainly as the sulfide mineral stibnite (Sb2S3). Antimony compounds have been known since ancient times and were powdered for use as medicine and cosmetics, often known by the Arabic name, kohl.

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nuclear energy is created through the process of splitting apart atoms and releasing large amounts of heat energy that can be co

shutvik [7]

Nuclear ﬁssion is a process by which the nucleus of an atom is split into two or more smaller nuclei, known as ﬁssion products. The ﬁssion of heavy elements is an exothermic reaction, and huge amounts of energy are released in the process.

3 0

2 years ago

When baking cookies, what is one chemical change that takes place?

Alexus [3.1K]

Answer:

A. chocolate chips melting

Explanation:

It is a chemical change. cookies will soften and melt slightly as the cookies bake. the change is solid to liquid.

6 0

2 years ago

Read 2 more answers

Which of the following is considered a complex (Macro)molecule?

Mazyrski [523]

Answer:

it is Glucose hope it helps

4 0

2 years ago

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Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol

max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

ΔG = ΔH - TΔS

On this note, it follows that;

ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

ΔG = 14.6 - 71.52
ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0

1 year ago

Read 2 more answers

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago