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2 years ago

Round this number to two significant figures. 634

Chemistry

2 answers:

Aneli [31]2 years ago

7 0

The answer to this is 630

loris [4]2 years ago

4 0

Answer : The correct answer is 630.

Explanation :

The following rules are used to round off a number to the required number of significant figures:

(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.

(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.

(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.

As we are given, 634

In the given number, there are 3 significant figures. Now we have to convert it into 2 significant figures.

According to the rules, round off the given measurement in 2 significant figures as 630 because last number i.e 4 is less than number 5 so, last digit become 0.

Therefore, the correct answer is 630.

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Ice and water constitute a system:

kondor19780726 [428]

Answer:

Explanation:

It's the same substance but in different states.

HETEROGENEOUS mixtures contain substances that are

not uniform in composition. The parts in the mixture can be separated by physical means.

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3 years ago

1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?

love history [14]

Answer:

$m=1.325gNa_2CO_3$

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

$eq-g=0.5eq-g/L*0.050L=0.025eq-g$

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

$mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3$

Finally, the mass is computed by using its molar mass (106 g/mol)

$m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3$

Regards.

7 0

3 years ago

Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans

loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

Answer: The pH of the solution is 10.4

Explanation:

We are given:

Molarity of sodium hypochlorite = 0.39 M

$pK_a$ of HClO = 7.50

We know that:

$pK_a=-\log K_a$

$K_a$ of HClO = $10^{-7.50}=3.16\times 10^{-8}$

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant = $3.16\times 10^{-8}$

$K_b$ = Base dissociation constant

Putting values in above equation, we get:

$10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}$

The chemical equation for the reaction of hypochlorite ion with water follows:

$ClO^-+H_2O\rightarrow HClO+OH^-$

Initial: 0.39

At eqllm: 0.39-x x x

The expression of $K_b$ for above equation follows:

$K_b=\frac{[HClO][OH^-]}{[ClO^-]}$

Putting values in above equation, we get:

$3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035$

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.00035M$

Putting values in above equation, we get:

$pOH=-\log (0.00035)=3.6$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.6=10.4$

Hence, the pH of the solution is 10.4

3 0

3 years ago

Water will move from a ____________ salt solution to a ____________ salt solution when they are across a differentially permeabl

const2013 [10]

The answers are low concentrated (dilute) and high concentrated respectively.

As the low concentrated salt solution has a higher water potential than that of the high concentrated salt solution, water molecules will flow from the region of higher water potential to the region of lower water potential, thus from the dilute salt solution to the high concentrated salt solution. This is due to the movement called osmosis. Note that osmosis also requires water to flow through a differentially permeable membrane, which means the membrane can allow certain substances (not all) to go in or out. If the differentially permeable membrane is not present, the movement of water molecules may be regarded as diffusion.

Therefore, the answers for the blanks are low concentrated and high concentrated.

6 0

3 years ago

In laboratory experiment, a NOVDEC Student was

Ede4ka [16]

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

The volume of the glucose solution to be prepared = 500 $cm^3$

Molarity of the glucose solution to be prepared = 1 M

i. Molar mass of glucose ( $C_1_2H_6O_6$ ) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii. mole = molarity x volume. Hence;

amount (in moles) of the glucose solution to be prepared

= 1 x 500/1000 = 0.5 mole

3 0

2 years ago