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My name is Ann [436]

4 years ago

6

Rn-222 has a half-life of 3.82 days. If 25.0 g of Radon-22 was originally present, approximately how many grams would be left af

ter 15 days?

1 answer:

horsena [70]4 years ago

6 0

Answer:

Approximately 0.39 g or 0.4 g if you're rounding up

Explanation:

15/3.82 = 3.92

Let's round that up to 4

That means 15 days is around 4 half lives

4 half lives means 1/16 of the original mass will be left

25/16 = 0.390625

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All the following ions have the same charge except Group of answer choices oxide. monohydrogen phosphate. peroxide. permanganate

Triss [41]

All the given species have the same charge except peroxide.

<h3>Charges on ions</h3>

The charge on monohydrogen phosphate is negative because of its formula, $[HPO4]^{2-}$ .

The charge on permanganate is negative. The formula is $MnO^-$

The charge of oxide ( $O^{2-}$ ) is negative.

The charge on peroxide is ( R−O−O−R) is neutral.

The charge on oxalate, $C2O4^{2-}$ , is negative.

Thus, only peroxide is unique.

More on charges on ions can be found here: brainly.com/question/11938054

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4 0

2 years ago

Enter the balanced complete ionic equation for HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq). Express your answer as a chemical equati

Marat540 [252]

<u>Answer:</u> The net ionic equation is given below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

These ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium carbonate is given as:

$2HCl(aq.)+K_2CO_3(aq.)\rightarrow H_2O(l)+CO_2(g)+2KCl(aq.)$

Ionic form of the above equation follows:

$2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+CO_3^{2-}(aq.)\rightarrow CO_2(g)+H_2O(l)+2K^+(aq.)+2Cl^-(aq.)$

As, potassium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$2H^+(aq.)+CO_3^{2-}(aq.)\rightarrow CO_2(g)+H_2O(l)$

Hence, the net ionic equation is given above.

7 0

3 years ago

Can someone help me please

Aloiza [94]

<span>A cloud collapses to form a star and disk. Planets form from this disk.

According to our current understanding, a star and its planets form out of

a collapsing cloud of dust and gas within a larger cloud called a nebula
.his dense, hot core becomes the kernel of a new star.

Hoped this helped :D
</span>

7 0

3 years ago

For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)

dangina [55]

Answer:

14.4 g of $CO_{2}$ can be produced.

Explanation:

Balanced equation: $2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O$

Molar mass (g/mol)

$C_{4}H_{10}$ 58.12

$O_{2}$ 32

$CO_{2}$ 44.01

So, 9.60 g of $C_{4}H_{10}$ = $\frac{9.60}{58.12}mol=0.165mol$

17.0 g of $O_{2}$ = $\frac{17.0}{32}mol=0.531mol$

According to balanced equation-

2 moles of $C_{4}H_{10}$ produce 8 moles of $CO_{2}$

So, 0.165 moles of $C_{4}H_{10}$ produce $(\frac{8}{2}\times 0.165)mol$ of $CO_{2}$ or 0.660 moles of $CO_{2}$

13 moles of $O_{2}$ produce 8 moles of $CO_{2}$

So, 0.531 moles of $O_{2}$ produce $(\frac{8}{13}\times 0.531)moles$ of $CO_{2}$ or 0.327 moles of $CO_{2}$

As least number of moles of $CO_{2}$ are produced from $O_{2}$ therefore $O_{2}$ is the limiting reagent.

So, maximum amount of $CO_{2}$ that can be formed = 0.327 moles

= $(0.327\times 44.01)g$

= 14.4 g

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3 years ago

What is light ?

Anuta_ua [19.1K]

Im pretty sure answer is c :)

5 0

3 years ago

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