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3 years ago

9.56 x 10^5 cg is how many grams?

Chemistry

1 answer:

Liula [17]3 years ago

4 0

Answer:

9.56×10¯⁷ g.

Explanation:

Mass in cg = 9.56×10¯⁵ cg

Mass in g =.?

Thus, we can convert 9.56×10¯⁵ cg to grams (g) as illustrated below:

Recall:

1 cg = 0.01 g

Therefore,

9.56×10¯⁵ cg = 9.56×10¯⁵ cg × 0.01 g / 1 cg

9.56×10¯⁵ cg = 9.56×10¯⁷ g

Therefore, 9.56×10¯⁵ cg is equivalent to 9.56×10¯⁷ g.

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MrRa [10]

<span>boron trichloride + water → boric acid + hydrochloric acid</span>

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What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

djverab [1.8K]

The empirical formula of the compound is CHCl₃.

<h3>Calculation:</h3>

Given,

Mass of carbon = 5.03 g

Mass of hydrogen = 0.42 g

Mass of chlorine = 44.5 g

Molecular weight of carbon = 12 g

Molecular weight of hydrogen = 1 g

Molecular weight of chlorine = 35.4 g

First, calculate the moles of each element,

Moles = given mass/ molecular weight

Moles of carbon = 5.03/12 = 0.42

Moles of hydrogen = 0.42/1 = 0.42

Moles of chlorine = 44.5/ 35.4 = 1.26

Divide the moles of each element by the smallest number of moles,

0.42 mol of C/ 0.42 = 1 C

0.42 mol of H/ 0.42 = 1 H

1.26 mol of Cl/0.42 = 3 Cl

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2 years ago

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

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