What type of symbolic relationship do a bat and cow have ?

You find a ring with a mass of 107g . You fill a graduated cylinder up with 10ml of water and put the ring into the cylinder . T

Korvikt [17]

Answer:

5 mL

Explanation:

Given data:

mass of ring = 107 g

volume of water = 10 mL

increase in volume = 15 mL

How much water displace = ?

Solution:

V (ring) = V (water + ring) - V (water)

V (ring) = 15 mL - 10 mL

V (ring) = 5 mL

when the ring is put into cylinder, volume is increased by 15 mL. The volume of water was 10 mL so water is displaced by 5 mL and the volume 5mL is the voulme of ring.

3 0

3 years ago

No. Of particles in H2SO4 in which 8grams oxygen is present????

steposvetlana [31]

Answer:

7?

Explanation:

Its somewhat hard to comprehend the question, but if the way I read it was right, its 7.

6 0

2 years ago

Is rock solution because it is made up of minerals, water, stone, and dirt?

julsineya [31]

Rocks are made up deposited minerals that form and condense into rocks

7 0

3 years ago

Read 2 more answers

Why doesn’t the KHP concentration have to be exactly 0.100M? Explain using complete sentences.

ollegr [7]

Answer:

The answer is in the explanation.

Explanation:

The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.

The reaction that explain this use is:

KHP + NaOH → KNaP + H2O

<em>where 1 mole of KHP reacts per mole of NaOH</em>

That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added

With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.

The moles of KHP are obtained from the volume and the concentration as follows:

Volume(L)*Concentration (Molarity,M) = moles of KHP

If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.

8 0

2 years ago

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

2 years ago