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3 years ago

Give the number of significant figures: 92.06 kg

Chemistry

2 answers:

Oksana_A [137]3 years ago

7 0

Answer: 4 significant figures

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.

All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.

All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.

All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.

Thus 92.06 has 4 significant figures.

Viefleur [7K]3 years ago

3 0

The number of significant figures is: 4

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac

valentinak56 [21]

Answer : The value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

First we have to calculate the $\Delta G_^o$ .

Formula used :

$\Delta G^o=-RT\times \ln K_p$

Now put all the given values in this formula, we get:

$\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})$

$\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole$

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}$

$Q=8.1\times 10^{5}$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})$

$\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole$

Therefore, the value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

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3 years ago

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3 years ago

Ca(s) + O₂(g) + S(s) → CaSO4(s)

san4es73 [151]

The mass of Calcium required to complete this reaction is 4.008 g.

Law of conservation of mass states that In a closed system, mass cannot be produced or destroyed, but it can be changed from one form to another.
The mass of the chemical constituents before a chemical reaction is equal to the mass of the constituents after the reaction.
In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.

In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.

So,

mass of Ca + mass of O₂ + mass of S = mass of CaSO4

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mass of Ca = 13.614 - 9.606 = 4.008 g

Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.

Learn more about mass conservation here:

brainly.com/question/2030891

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2 years ago