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Anna11 [10]
2 years ago
6

If it take 118j of heat to increase the temperature of a 5.0g block of silver by 15 c, what is the specific heat capacity

Chemistry
1 answer:
KatRina [158]2 years ago
7 0
Q=mc(delta)T
118=0.005×c×15
c=118/(0.005×15)

c= 1573 J/kg/C
You might be interested in
Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

3 0
2 years ago
2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b
Virty [35]

Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: 2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl

     Compound                                 Molar mass (g/mol)

         NaOH                                           39.997

         MgCl_{2}                                           95.211

        Mg(OH)_{2}                                        58.3197

So, 2.50 g of NaOH = \frac{2.50}{39.997} mol of NaOH = 0.0625 mol of NaOH

      4.30 g of MgCl_{2}  = \frac{4.30}{95.211} mol of MgCl_{2} = 0.0452 mol of MgCl_{2}

According to balanced equation-

2 mol of NaOH produce 1 mol of Mg(OH)_{2}    

So, 0.0625 mol of NaOH produce (\frac{0.0625}{2}) mol of NaOH or 0.03125 mol of NaOH

1 mol of MgCl_{2} produces 1 mol of Mg(OH)_{2}

So, 0.0452 mol of MgCl_{2} produce 0.0452 mol of Mg(OH)_{2}

As least number of moles of Mg(OH)_{2} are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of Mg(OH)_{2} would be produced = 0.03125 mol

                                                                           = (0.03125\times 58.3197) g

                                                                           = 1.822 g

6 0
2 years ago
The graph shows the amount of a gaseous product formed over time during two trials of a reaction. A different concentration of a
ss7ja [257]

Answer:

Trial 2, because the amount of product formed per unit time is higher.

Explanation:

Message me for extra help.

8 0
2 years ago
Read 2 more answers
The bonds in the compound MgSO4 can be described as
Butoxors [25]
C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond
7 0
2 years ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

4 0
2 years ago
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