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2 years ago

Scientists used a powerful telescope to discover the caloris crater on mercury

Chemistry

1 answer:

Drupady [299]2 years ago

6 0

Answer:

True

Explanation:

Mercury is one of the five classical planets visible with the N4KED eye and is named after the swift-footed Roman messenger god. It is not known exactly when the planet was first discovered - although it was first observed through telescopes in the seventeenth century by astronomers Galileo Galilei and Thomas Harriot.

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For the following reaction, 28.6 grams of zinc oxide are allowed to react with 9.54 grams of water . zinc oxide(s) water(l) zinc

maw [93]

Answer:

34.9 g of Zn(OH)₂ is the maximum mass that can be formed

Explanation:

Let's state the reaction:

ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)

First of all, we need to determine the moles of each reactant and state the limiting:

28.6 g . 1mol /81.38 g = 0.351 moles of ZnO

9.54 g . 1mol /18 g = 0.53 moles of water

As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.

Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:

0.351 mol . 99.4 g /1mol = 34.9 g

3 0

3 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

According to the collision theory of chemical reactions, an increase in the number of effective reactant collisions per unit tim

olasank [31]

1. true, more effective collisions per second. Faster reaction
2. <span>How can the reaction be slowed down? (talking about how fast or slow the reaction is)
</span>3. True
4. Rate lower if surface area decrease
5. Fine powder form
6. True
7. False
Image questions:
1. False
2. Energy of reactants higher than products.

7 0

3 years ago

Read 2 more answers

A student was asked to name an element that reacts like chlorine (Cl). Which element should the student name?

kramer

Answer:

Bromine (BR2)

Explanation:

Cl2 + Br2 = ClBr

5 0

3 years ago

What process would best separate a solution of liquid water and liquid rubbing alcohol