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3 years ago

Five elements named after color

1 answer:

4 0

Caesium -bluish(Latin) Chlorine -yellow/green (Greek) Iodine -violet (Greek) <span>Rhodium -rose (Greek) Sulphur - yellow (Arabic)</span>

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A balloon sits in the sunlight causing it to heat up and explode. Which of the following

luda_lava [24]

I think it’s A, the particles of gas inside the ballon move faster and decrease pressure in

6 0

2 years ago

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

4 years ago

If the density of gold is 19.3 g/cm3 what would be the volume of 550 g of gold?

RUDIKE [14]

Answer:

28.497 cm3

Explanation:

Formula

D=m/v

Given data:

density = 19.3g/cm3

mass = 550 g

Now we will put the values in formula:

V=m/d

V=550 g/ 19.3 g/cm3 = 28.497 cm3

So the volume of gold is 28.497 cm3.

7 0

4 years ago

1. A charged group of covalently bonded atoms is known as

katrin2010 [14]

Its known as covalently bonded atoms

7 0

3 years ago

Help please thank you

Vikki [24]

Answer:

B I think I am pretty sure

8 0

3 years ago