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3 years ago

Convert the following into scientific notation 0.8441

Chemistry

1 answer:

ASHA 777 [7]3 years ago

6 0

Answer:

the answer is D

hope this will help ❤️

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What happens when ethylene is subjected to ozonolysis?

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When ethylene is subjected to ozone it forms Acetaldehyde and oxygen is released

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2 years ago

Select The TWO answers that are correct, Put trust in your answer brainliest

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B and C

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3 years ago

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Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li (s) + N2 (g) → 2Li3N (s) In a particular ex

seraphim [82]

Answer: 6.71 g

Explanation: $6Li(s)+N_2(g)\rightarrow 2Li_3N$

${\text{no of moles}}=\frac{\text{Given mass}}{\text{Molar mass}}$

${\text {moles of lithium}}=\frac{4g}{6.914g/mol}=0.578moles$

$\text{moles of nitrogen}=\frac{4g}{28g/mol}=0.143moles$

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus lithium is the limiting reagent and nitrogen is the excess reagent.

As can be seen from the balanced chemical equation, 6 moles of lithium reacts with 1 mole of nitrogen to give 2 moles of lithium nitride.

Thus 0.578 moles of lithium react with 0.096 moles of nitrogen.

6 moles of lithium give = 2 moles of lithium nitride

Thus 0.578 moles of lithium give= $\frac{2}{6}\times {0.578}=0.19moles$ of lithium nitride.

Mass of lithium nitride $Li_3N={\text {no of moles}}\times {\text {Molecular mass}}$

Mass of lithium nitride $Li_3N$ = ${0.192moles}\times {34.83g/mol}=6.71g$

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3 years ago

A plant grows into a shrub a physical or chemical change?<br>Explain why?

kherson [118]

A plant growing into a shrub is an example of a chemical change because once the plant is in its shrub form you can't revert it back.

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3 years ago

A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

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3 years ago