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2 years ago

What is the concentration of a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7

1 answer:

5 0

SrSo4 = Sr(2+) + SO4(2-)

Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).

When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4

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A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

GrogVix [38]

Answer:

The answer to the question is

Suki is 1.12 m close to the end of the steel beam before the beam begins to tip

Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

or x = -0.881 m on the other side of the support

Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

7 0

2 years ago

Coffee is<br><br> homogeneous.<br><br> heterogeneous.

Basile [38]

The answer is homogeneous

3 0

3 years ago

PLEASE HELP

torisob [31]

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4 0

3 years ago

quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe

12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

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1 year ago

Mass=4grams Density=2g/ml

Tom [10]

Answer:

you can also get the target audience for the first one in the world 35ml.

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3 years ago