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bezimeni [28]
4 years ago
6

If conducting research for a paper on restricting cell phone usage in automobiles, which source would be most effective to use i

n your argument?
Chemistry
2 answers:
Damm [24]4 years ago
6 0

Answer: A) a recent court case

My name is Ann [436]4 years ago
4 0
 A source you could use is a statistic from a car company of cell phone usage inside a moving vehicle compared to accidents occurring due to it.
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Name two other than chlorine (CI) that are likely to react with sodium (Na)
agasfer [191]
F and O
because they are more reactive than Cl
4 0
3 years ago
What is the chemical formula of magnesium bromide? A. MgBr2 B. MgBr C. Mg2Br2 D. Mg2Br
DanielleElmas [232]
A. MgBr2
because Mg2+ and Br-
7 0
3 years ago
Give 10 chemical properties of common polymers​
algol [13]

Answer:

Some of the useful properties of various engineering polymers are high strength or modulus to weight ratios (light weight but comparatively stiff and strong), toughness, resilience, resistance to corrosion, lack of conductivity (heat and electrical), color, transparency, processing, and low cost

Explanation:

3 0
3 years ago
Read 2 more answers
Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)
vampirchik [111]

Answer: Oxygen is the limiting reagent.

Explanation: CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)

As can be seen from the given balanced equation:

3 moles of O_2 reacts with 1 mole of CS_2

1.52 moles of O_2 reacts with=\frac{1}{3}\times 1.52=0.51moles of CS_2

Thus O_2 is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of CS_2 will remain as such and thus  CS_2 is an excess reagent.

6 0
3 years ago
Calculate the bond energy of HBr.Bond energy of H2 is 436KJ/mol and that of Br2 is 193KJ/mol.
lina2011 [118]

Answer:

ΔH3 =  1/2 (629) - ΔH^0

Explanation:

Given data:

Bond energy of H2 = ΔH1 =  436 Kj/mol

Bond energy of Br2 = ΔH2 = 193 Kj/mol

To find:

Let bond energy of HBr = ΔH3 = ?

Equation:

H2 + Br2 → 2HBr

enthalpy of formation of HBr = ΔH1 + ΔH3 - 2(ΔH3)

ΔH^0 = 436 + 193 -   2(ΔH3)

(436 + 193) - ΔH^0 = 2(ΔH3)

ΔH3 =  1/2 (629) - ΔH^0

3 0
4 years ago
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