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3 years ago

Which element was first discovered in space before it was discovered on Earth?

2 answers:

8 0

Helllllooo Friend __________________________:-) :-):-) :-)

©©©©©here is your answer©©©©©

Helium is named from the Greek helios, "the Sun", where the element was discovered through spectroscopy.

Hope this answer help you!!!!!!!!!!! :-) :-) :-) :-) :-) :-)

faltersainse [42]3 years ago

7 0

Helium was first discovered in space before it was on earth

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At STP, which substance has metallic bonding?

kogti [31]

Silver has metallic bonding.

Silver is a very typical and main metal. The negatively charged electrons distribute themselves throughout the entire piece of metal and form non directional bonds between the positive silver ions, which is metallic bonding, and what silver contributes.

7 0

4 years ago

Read 2 more answers

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

6 0

3 years ago

many cars are fitted with airbags which inflate in an accident. Airbags contain solid sodium azide, NaN₃, which decomposes rapid

Lesechka [4]

The initial balanced equation of decomposition is:

2NaN3 —> 2Na + 3N2

Since 1 mole of a gas occupies 24dm^3 at room temperature and pressure:

72dm^3 / 24dm^3 = 3 moles of Nitrogen which required 2 moles of Sodium Azide.

Since moles = mass/Relative molecular mass or molar mass

2 moles x 65~ g/mol = 130g

For the empirical formula:

Element H N
% Mass 2.33% 97.7%
Mass/Mr 2.33 6.97
2.33 7~

Divide by lowest number

1 3

Therefore empirical formula is

H1N3

5 0

4 years ago

10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo

Natalija [7]

Answer:

$\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}$

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

$\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}$

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

$\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}$

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.150 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.150 - x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150$

(ii) Solve for x

$\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}$

(iii) Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}$

(c) pH at equivalence point

(i) Calculate the moles of each species

$\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}$

B + HI ⇌ BH⁺ + I⁻

I/mol: 3.00 3.00 0

C/mol: -3.00 -3.00 +3.00

E/mol/: 0 0 3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

$\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}$

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

BH⁺+ H₂O ⇌ H₃O⁺ + B

I/mol·L⁻¹: 0.0750 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.0750 - x x x

$K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}$

$\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}$

(iv) Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}$

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

8 0

3 years ago

The equation below represents the combination of gaseous atoms of non-metal X and of hydrogen to form gaseous X2He6 molecules. 2

vichka [17]

Answer:

D +405.0kJ mol-¹

Explanation:

Since bond energy is the energy required to break a bond, the energy of dissociation of X₂H₆ = +2775 kJmol⁻¹.

Since there is one X-X bond and six X-H bonds,

Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆.

Since bond energy of one X-H bond = 395 kJ mol⁻¹, then

Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆

Bond energy of one X-X bond + 6 × one X-H bond = +2775 kJmol⁻¹.

Bond energy of one X-X bond + 6 × 395 kJ mol⁻¹ = +2775 kJmol⁻¹.

Bond energy of one X-X bond + 2370 kJ mol⁻¹ = 2775 kJmol⁻¹

Bond energy of one X-X bond = 2775 kJmol⁻¹ - 2370 kJ mol⁻¹

Bond energy of one X-X bond = +405 kJmol⁻¹

6 0

3 years ago