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3 years ago

What is the significant digit in 23.45

1 answer:

7 0

If you mean the number of Significant Figures/Digits in 23.45, it would be 4. This is because every single non-zero digit is counted as a significant figure

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Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283

Lyrx [107]

First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2

6 0

3 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0

3 years ago

A prion is an infectious self-reproducing protein structure. true or false

Alexandra [31]

Answer:

True is the correct answer.

Explanation:

The statement that a prion is an infectious, self-reproducing protein structure is true.

prion is an infectious particle and they do not have genetic material.

Prion present in the brain region that results in deadly neurodegenerative illnesses in humans and animals.

Prions are the self-producing proteinaceous infectious capable of transferring infection in the absence of the nucleic acids.
prions enter in the brain by an infection and they emerge from the variation in the gene that encodes the protein and once the prions present in the brain multiply by causing the benign proteins to refold into an abnormal form.

8 0

3 years ago

Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O

Bess [88]

40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O

5 0

3 years ago

Although both N2 and 02 are naturally present in the air we breathe, high levels of NO and NO2 in the atmosphere occur mainly in

joja [24]

Answer:

(a) Increasing the temperature adds heat which is a reactant shifting the equilibrium rightwards.

(b) Pressure has no effect since the change in the number of moles is zero.

Explanation:

Hello,

In this case, one could represent the given reaction as:

$N_2(g)+O_2(g)+ \Delta _rH \leftrightarrow 2NO$

Since it is endothermic. Thus, solving the (a) statement, one identifies the heat as a reagent, that is why the reaction cools down as it progress, therefore, by increasing the temperature, heat is added, that is, a reagent is added, which shifts the equilibrium rightwards, in other words, more NO is produced so its concentration increases.

Furthermore, for the (b) statement, since the change in the number of moles is zero, based on the stoichiometric coefficients as shown below:

$\Delta \nu =2-1-1=0$

Such value implies that the pressure has no effect on the concentration, taking into account the following form of the law of mass action:

$Kp=Kc(RT)^{\Delta \nu }$

Thus, since $\Delta \nu =0$ , $Kp=Kc$ , so no effect in concentration is due to the pressure.

Best regards.

6 0

3 years ago