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Pepsi [2]
3 years ago
7

Complete the table to summarize the properties of the different subatomic particles by typing in 'yes"

Chemistry
2 answers:
Free_Kalibri [48]3 years ago
3 0
The answer:
A) Yes
B)Yes
blsea [12.9K]3 years ago
3 0

Answer:

A=yes

B=yes

Explanation:

Just did the same thing

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Determine whether the given compound name or formula contains a polyatomic ion. N H 4 C l Choose... Magnesium sulfate Choose...
Morgarella [4.7K]

Answer:

NH₄Cl, Magnesium sulfate, KCN

Explanation:

Determine whether the given compound name or formula contains a polyatomic ion.

  • NH₄Cl. YES. It contains the polyatomic ion ammonium NH₄⁺.
  • Magnesium sulfate. YES.  It contains the polyatomic ion sulfate SO₄²⁻.
  • Sodium phosphide Na₃P. NO.
  • SO₃. NO.
  • Calcium hydroxide Ca(OH)₂. NO.
  • KCN. YES. It contains the polyatomic ion cyanide CN⁻.
4 0
3 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

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
4 years ago
How many moles of CO2 are produced from the combustion of 5.25 moles of CH3OH?
iVinArrow [24]

First, we write the reaction for CH3OH combustion

CH3OH+3/2O2--->CO2+2H2O

for 1 mole of methanol, we get 1 mole of CO2, therefore for 5,25 moles of methanol we will get 5,25 moles of CO2

4 0
3 years ago
Forecasting the weather for a specific area is relatively easy because of all the technology that is available.
AleksAgata [21]

Answer:

A

Explanation:

4 0
4 years ago
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