Can someone please help me with number 10

Transition metals often form ions with a charge of _____-

Bess [88]

+1 for transition metals

5 0

2 years ago

Please help me i’m in need!

makkiz [27]

Answer:

balanced............

3 0

2 years ago

What characteristics do the elements in a group share? In a period, share?

Katyanochek1 [597]

Answer:

The elements in same period have same principle quantum number or energy shell.

The elements in same group shows similar chemical and properties.

Explanation:

Inn group:

The elements in same group i.e present in vertical column shows similar chemical properties.

The elements in same group having same number of valance electrons. while in chemical reaction bonds are break and formed and valance electrons are involved. That's why elements in same group having same number of valance electrons and shows similar chemical properties.

In period:

While as we move from left to right the number of valance electron increase by one in every element. But the electron is added in same shell which means that their physical and chemical properties are different but principal quantum number is same.

3 0

2 years ago

The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con

Aloiza [94]

Use the formula for second order reaction:

$\frac{1}{C} = \frac{1}{C_0} + kt$

C = concentration at time t
C0 = initial conc.
k = rate constant
t = time

1st equation : $\frac{1}{0.56} = \frac{1}{C_0} + 115k$

2nd Equation: $\frac{1}{0.033} = \frac{1}{C_0} + 735k$

Find $\frac{1}{C_0}$ from 1st equation and put it in 2nd equation:

$\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k$

$\frac{1}{0.033} - \frac{1}{0.56} = 620k$

k = 0.046

3 0

3 years ago

Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a

Annette [7]

<h3>solution:</h3>

The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:

s= 8.2652 kJ/kgK

The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:

T₂ = T₁+ T₂ - T₁/ 8₂ - 8₁ ( s - s₁)

= (400 + 500 - 400/8.3271 - 8.0347 (8.2652 - 8))

= 478.83°C

The final enthalpy is determined in the same way:

h₂= h₁ + h₂ - h₁/s₂ - s₁ ( s - s₁)

= ( 3275.5+ 3486.6 - 3275.5/ 8.3271 - 8.0347) (8.265)

= 3441.91 kJ/kg

6 0

1 year ago