All categories

3 years ago

Hi can you pls help :)

Chemistry

1 answer:

Luda [366]3 years ago

8 0

Answer:i think it is c, it explodes in water because that’s a chemical reaction

Explanation:

You might be interested in

In what unit is formula mass expressed?

Illusion [34]

Italicized lowercase m

8 0

4 years ago

For the following reaction, provide the missing information

jenyasd209 [6]

Answer:

19. Option B. ⁰₋₁B

20. Option D. ²¹⁰₈₄Po

Explanation:

19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

Thus, we can determine ʸₓZ as follow:

228 = 228 + y

Collect like terms

228 – 228 = y

y = 0

88 = 89 + x

Collect like terms

88 – 89 = x

x = –1

Thus,

ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B

20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

Thus, we can determine ᵘᵥX as follow:

u = 206 + 4

u = 210

v = 82 + 2

v = 84

Thus,

ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po

ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He

8 0

3 years ago

Please help me I need these answers

Vladimir [108]

Answer:

i am rrly bad at chemistry srry fffdfhdhdgduehgfhfhdhhfhdhdhdhdhdudhd

7 0

3 years ago

Nitroglycerine decomposes violently according to the chemical equation below. What mass of carbon dioxide gas is produced from t

Ulleksa [173]

Explanation:

Below is an attachment containing the solution.

5 0

4 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

3 years ago