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3 years ago

3 & 4 Help pleaseeeee

Chemistry

1 answer:

kvasek [131]3 years ago

4 0

3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...

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CHEMISTRY HELP<br> Once the following equation is balanced, what is the correct coefficient for Z₂?

yawa3891 [41]

Answer:

The coefficient of Z₂ is 1.

Explanation:

From the question given above:

X + ZY —> XY + Z₂

Next, we shall balance the equation to obtain the coefficient of Z₂. This can be obtained as follow:

X + ZY —> XY + Z₂

There is 1 atom of Z on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of ZY as shown below:

X + 2ZY —> XY + Z₂

There are 2 atoms of Y on the left side and 1 atom on the right side. It can be balance by putting 2 in front of XY as shown below:

X + 2ZY —> 2XY + Z₂

Now, we have 1 atom of X on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of X as shown below:

2X + 2ZY —> 2XY + Z₂

Now the equation is balanced.

Thus, the coefficient of Z₂ is 1.

8 0

2 years ago

As a human being, we are considered to be a ..... because we rely on other organism for food

ryzh [129]

Heterotrophs i think :)

3 0

3 years ago

Read 2 more answers

The gaseous product of a reaction is collected in a 25.0L container at 27.0 C. The pressure in the container is 3.0atm and the g

NeX [460]

Answer: The molar mass of the gas is 31.6 g/mol

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 3.0 atm

V = Volume of gas = 25.0 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $27.0^0C=(27.0+273)K=300K$

$n=\frac{PV}{RT}$

$n=\frac{3.0atm\times 25.0L}{0.0821 L atm/K mol\times 300K}=3.04moles$

Moles = $\frac{\text {given mass}}{\text {Molar mass}}$

$3.04=\frac{96.0g}{\text {Molar mass}}$

${\text {Molar mass}}=31.6g/mol$

The molar mass of the gas is 31.6 g/mol

4 0

2 years ago

A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the

Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

$\% M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\% =\frac{m_M}{m_M+32.00}*100 \%$

Thus, we solve for the molar mass of the metal to obtain:

$59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol$

For the subsequent problems, we proceed as follows:

$4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2$

$0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH$

$0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH$

Regards!

7 0

2 years ago

Which sphere would this be found in?

andre [41]

That is correct…….. i think

8 0

2 years ago

Read 2 more answers