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2 years ago

8

How many grams of barium sulfate are produced if 125 mL of .050 M sodium sulfate are reacted with excess barium nitrate in the f

ollowing reaction?
Ba(NO3)2 + Na2SO4 = BaSO4 + NaNO3

1 answer:

katrin [286]2 years ago

8 0

Answer:

2.79g

Explanation:

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2 HCI → H2 + Cl2<br> Endothermic or exothermic

Oksana_A [137]

Answer:

Exothermic

Explanation:

I think it is exothermic because heat is being released

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2 years ago

An atom has completely filled inner shells and a single valence electron in an excited p state. The filled inner shells have an

umka2103 [35]

Answer:

1

Explanation:

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l=-1: theta=135 deg

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2 years ago

Who wants a metal?

tiny-mole [99]

Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

A. Both have the same molar mass.

B. Both have the same number of ions.

C. Both are made up of 6.02 mc014-1 1023 molecules.

D. Both are made up of 6.02 mc014-2 1023 formula units.

The correct answer on E.D.G is ---Both are made up of 6.02 mc014-2 1023 formula units. D

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3 years ago

Read 2 more answers

A chemistry student needs 10.00 g of isopropenylbenzene for an experiment. He has available 120. g of a 42.7% w/w solution of is

Vanyuwa [196]

<u>Answer:</u> The mass of solution that the chemistry student should use is 23.4 grams

<u>Explanation:</u>

We are given:

Available mass of isopropenylbenzene = 120. g

Amount of isopropenylbenzene needed by chemistry student = 10.00 g

42.7 % (w/w) solution of isopropenylbenzene.

This means that 42.7 grams of isopropenylbenzene is present in 100 grams of solution.

To calculate the mass of solution for given needed of isopropenylbenzene, we apply unitary method:

For 42.7 grams of isopropenylbenzene, the amount of solution needed is 100 grams

So, for 10.00 grams of isopropenylbenzene, the amount of solution needed will be = $\frac{100}{42.7}\times 10=23.4g$

Hence, the mass of solution that the chemistry student should use is 23.4 grams

3 0

2 years ago

9. Thallium-208 has a half-life of 3.053 min. How long will it take for 120 g of it to decay

damaskus [11]

Answer:

12.213 minutes will be taken for 120 g-Thalium-208 to decay to 75 grams.

Explanation:

Radioactive isotopes decay exponentially in time, the mass of the isotope ( $m(t)$ ), in grams, is described by the formula in time ( $t$ ), in minutes:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

Where:

$m_{o}$ - Initial mass of the isotope, in grams.

$\tau$ - Time constant, in minutes.

In addition, the time constant associated with the isotope decay can be described in terms of half-life ( $t_{1/2}$ ), in minutes:

$\tau = \frac{t_{1/2}}{\ln 2}$ (2)

If we know that $m(t) = 7.5\,g$ , $m_{o} = 120\,g$ and $t_{1/2} = 3.053\,min$ , then the time taken by the isotope is:

$\tau = \frac{t_{1/2}}{\ln 2}$

$\tau = \frac{3.053\,min}{\ln 2}$

$\tau \approx 4.405\,min$

$t = -\tau \cdot \ln \frac{m(t)}{m_{o}}$

$t = -(4.405\,min)\cdot \ln \left(\frac{7.5\,g}{120\,g} \right)$

$t \approx 12.213\,min$

12.213 minutes will be taken for 120 g-Thalium-208 to decay to 75 grams.

4 0

2 years ago