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2 years ago

7

What is the barometric trend of the weather data?

1 answer:

larisa86 [58]2 years ago

4 0

Answer:

the change of atmospheric pressure during the last few (generally three) hours before a regular observation.

Explanation:

Even small changes in barometric pressure can greatly impact the weather. The Weather Trend directional arrow icon, included on many AcuRite weather station products, will show one of the following three barometric pressure trend directions, indicating potential future weather conditions.

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You are given 100 g of a compound. The compound is composed of 37% hydrogen and 63% oxygen. How many grams oxygen are present in

USPshnik [31]

$mass_{oxygen} = \%mass_{oxygen} \times mass_{compound} \\ m = 63\%(100) \\ m = 0.63(100) \\ m = 63 \: grams$

<h2>Option d</h2>

7 0

1 year ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

3 years ago

Can u help me with this

dolphi86 [110]

Answer:

c

Explanation:

the reason is because your a homosiejenenddj

8 0

2 years ago

The below given compound is slightly soluble in water.

nordsb [41]

Answer:

an alkali

Explanation:

an alkali is a base that is slightly soluble in water

4 0

2 years ago