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3 years ago

5. Which coefficients correctly balance the formula equation

Chemistry

1 answer:

Tju [1.3M]3 years ago

8 0

Answer:

which of these sentences contains a correctly formed possessive form of the name andres

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Write the equation for the reaction described: A solid metal oxide, Ag2O, and hydrogen are the products of the reaction between

Harlamova29_29 [7]

Answer:

2Ag + H2O -----> Ag2O + 2H

Explanation:

2Ag + H2O -----> Ag2O + 2H is the equation of the reaction between metal and steam. Silver reacts with water (steam) forming silver oxide and hydrogen gas. When the metals react with steam it produces the solid metal oxide and hydrogen gas. On the surface o metals, a protective layer of aluminium oxide is formed that keeps water away from the metal so we can say that silver oxide and hydrogen are formed from the reaction of silver metal and steam.

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3 years ago

What is the dependent variable in the experiment shown

Ilia_Sergeevich [38]

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3 years ago

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What happens when you mix hydrogen and oxygen?

Lerok [7]

What happens is it makes water

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3 years ago

Urgent!! A chemist measured 5.2 g copper(II) bromide tetrahydrate (CuBr2•4(H2O)). How many moles were measured out? Answer in un

bezimeni [28]

The moles which were measured out is calculated using the following formula

moles = mass/molar mass

molar mass of CuBr2.4H20 = 63.5 Cu + ( 2 x79.9) br + ( 18 x4_) h20 = 295.3 g/mol

moles is therefore= 5.2 g/ 295.3 g/mol= 0.0176 moles

4 0

3 years ago

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How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0

3 years ago