All categories

3 years ago

Round to 4 significant figures. 35.5450

Chemistry

2 answers:

castortr0y [4]3 years ago

8 0

Answer:

35.55 to nearest 0.01 or

hundredth place

Lelu [443]3 years ago

3 0

Answer:
35.5450 will be rounded to 35.55
Explanation:
=35.5450
if the last digit is less than 5 then it will be ignored
=35.545
when the dropping digit is 5 then the retaining digit will increse by a factor of 1
=35.55
i hope this will help you

You might be interested in

Could someone please help me??

gizmo_the_mogwai [7]

7.9×10^-1 Kg/m³

or, 0.79 Kg/m³

Hope it helps you...

8 0

2 years ago

Read 2 more answers

How do air masses flow?

zepelin [54]

<em>How do air masses flow?</em>

<em>from regions of high pressure to low pressure</em>

In general, cold air masses tend to flow toward the equator and warm air masses tend to flow toward the poles.

6 0

2 years ago

How many attoms are equal 6.19 moles of hydrogen?

Wittaler [7]

Answer:

3.727x10^24 atoms.

Explanation:

When calculating how many molecules are in a compound or element, we refer to Avogardo's number.

Your equation will look like this: 1 mole/ 6.022x10^23 molecules (Avogardo's number).

Multiply 6.19 by Avogardo's number to get your answer.

4 0

2 years ago

What is the molarity of a solution that contains 1.78 moles of KNO3 dissolved in 2.50L of water?

Alexxandr [17]

Answer:

molarity= 0.712M

Explanation:

From n=CV

n=1.78, C=?, V= 2.5L

1.78=C×2.5

C= 0.712M

4 0

2 years ago

300 mL of 0.200 M Pb(NO3)2 is added to 200 mL of 0.0500 M NaCl(aq) at 25°C. Will a precipitate of PbCl2 form at 25 °C? Tip: You

Bas_tet [7]

Explanation:

It is given that volume of $Pb(NO_{3})_{2}$ is 300 mL and molarity is 0.200 M.

Volume of NaCl is 200 mL and molarity is 0.050 M.

The chemical reaction will be as follows.

$PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$

$K_{sp}$ for $PbCl_{2}$ is given as $1.7 \times 10^{-5}$ .

As, molarity is number of moles present in liter of solution.

Hence, moles of $Pb^{2+}$ (aq) will be calculated as follows.

moles of $Pb^{2+}$ (aq) = $0.300 L \times 0.200 M$

= 0.06 mol

$[Pb^{2+}]$ = $\frac{0.06 mol}{0.5 L}$

= 0.120 M

Mole of $Cl^{-}(aq)$ = $0.2 L \times 0.05 M$

= 0.010 M

Now, Q = $[Pb^{2+}][Cl^{-}]^{2}$

= $0.120 \times (0.010)^{2}$

= $1.2 \times 10^{-5}$

As, Q < $K_{sp}$ hence, there will be no formation of $PbCl_{2}$ precipitate.

3 0

2 years ago