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2 years ago

How many significant figures are in 0.00102?

Chemistry

2 answers:

Lunna [17]2 years ago

8 0

Answer:

3 digits

Explanation:

The first nonzero digit is counted as a significant digit and the zero inbetween 2 nonzero digits count too.

Therefore, the significant digits are 0.00102.

zaharov [31]2 years ago

7 0

Answer:

Explanation:

102 are significant figures

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1. power utilizes moving water to generate electricity.

Marianna [84]

Hydroelectric energy

8 0

3 years ago

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

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5 0

1 year ago

G.com what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume

tiny-mole [99]

7.379 * 10^(-4) is measured, hence prone to error, either human error or via measuring device. In this case,
100 cm = 1 m is written in stone and is unquestionable.
The density of the gold is 19.3 g/cm^3 and could be an approximation.
The approximation is good to at least one night.

5 0

3 years ago

How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?

Kitty [74]

Answer:

21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C7H14 + 21O2 —> 14CO2 + 14H2O

From the balanced equation above, 2L of C7H14 produced 14L of H2O.

Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.

Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2O will occupy

26.25L i.e

Xmol of H2O = 26.25/22.4

Xmol of H2O = 1.172 mole

Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

Number of mole H2O = 1.172 mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O =..?

Mass = mole x molar mass

Mass of H2O = 1.172 x 18

Mass of H2O = 21.10g

Therefore, 21.10g of H2O is produced from the reaction.

3 0

3 years ago

Solid carbon tetrachloride, CCl4(s), is represented by the diagram above. The attractions between the CCl4 molecules that hold t

Ne4ueva [31]

Hi, you've asked an incomplete question. Here's the diagram that completes the question.

Answer:

(B) nonpolar covalent bonds

Explanation:

This structure in the diagram rightly fits the description of a non-covalent bond because there is an equal sharing of electrons of Carbon (C) and Chlorine (Cl).

Remember too that these elements are in their solid-state, hence the CCl4 (carbon tetrachloride) molecules are held strongly together.

7 0

3 years ago