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3 years ago

Ca(OH) =CaO + H2O

Chemistry

1 answer:

DENIUS [597]3 years ago

8 0

Answer:

2 combustion

3 combustion

4 decomposition

5 replacement

6 Decomposition

Note:

Im not really sure if those r right.

It may be right actually.

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Iodine-131 is a radioactive material that plays a role in natural gas production and has a half-life of 8 days. A 320-mg sample

svetlana [45]

Time of decays approximately 2.853 days

<h3>Further explanation</h3>

Given

mass = 320mg iodine-131

No=320 mg

Nt=250 mg

t1/2 = 8 days

Required

t=time of decays

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

Input the value

$\tt 250=320.\dfrac{1}{2}^{t/8}\\\\log\dfrac{25}{30}=t/8.log~0.5\rightarrow t=2.853~days$

3 0

3 years ago

The substance being dissolved is known as the __________.

pishuonlain [190]

<span>a. solute. The solute is the substance usually dissolved in the solvent to form the solution.</span>

5 0

3 years ago

Read 2 more answers

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Write any five example of radical with their velencies

garri49 [273]

Answer:

Terms in this set (13)

Hydroxide. OH.

Nitrate. NO3.

Ammonium. NH4.

Bicarbonate/Hydrogen Carbonate. HCO3.

Bisulphate. HSO4.

Chlorate. ClO3.

<h2>hope it's correct</h2>

7 0

3 years ago

4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) → 4 NaAu(CN)2(aq) + 4 NaOH(aq) (2) 2 NaAu(CN)2(aq) + Zn(s) → 2 Au(s) +Na2[Zn(CN)4](aq) I

marysya [2.9K]

Answer:

72.57 grams

Explanation:

The mass percentage of gold in the ore is 0.19 %

The mass of ore used = 23 kg

The mass of gold in the ore = $\frac{0.19X23}{100}=0.0437kg$

moles of gold in the given mass of ore = $\frac{mass}{Atomic mass}=\frac{0.437X1000}{197}=2.22mol$

As per given equation four moles of Au is giving four moles of complex compound on reacting with NaCN

Then in second reaction two moles of the complex is reacting with one mole of Zinc

Thus two moles of gold are reacting with one mole of Zinc

The moles of Zinc needed = 0.5 X 2.22 mol = 1.11 moles

The mass of Zinc needed = moles X atomic mass =1.11 X 65.38 = 72.57 grams

6 0

3 years ago