Answer:
57 grams of H3PO4
Explanation:
M= moles/ liters
convert mL to L
234 mL x 1L/1000mL = 0.234L
Rearrange the Molarity formula to solve for moles.
moles= MxL
moles= 2.5M x 0.234L
moles= 0.585 mol
Use the molar mass of H3PO4 to get to grams
0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4
round to two sig figs for 57 grams
Answer:
2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O
Explanation:
This a neutralisation reaction. The first thing to note is that this type of reaction forms salt and water.
Note that the formula of water is H2O. So write this at the right hand side.
==> H2O
Next note that the salt should contain the acid radical (NO3) and the metallic part of the base (Ca) written well following the rule for writing chemical equation; thus
==> Ca(NO3)2 + H2O
The next step is to write the reactants at the right hand side of the equation;
HNO3 + Ca(OH)2 ==> Ca(NO3)2 + H2O
The final step is to count the number of atoms in each side of the equation to make sure they are equal and balanced.
In the above equation, H atoms are three at the left hand side add 2 to to H- containig compound on both side as to balance up. Thus
2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O
The equation 2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O is balance.
Answer: The general formulae for moles is n=m/mr so now we have being given to find the mass so all we have to do is to change subject
Explanation: so we have to change subject in this question to m= n× mr . so in the question below we have being given the mole as 1.5mol/dm³ so all we have to do is to find the molecular relative mass(mr) .
to find the molecular relative mass of sodium hydroxide (NAOH) we add all of the atomic masses of all the atoms present so here we have sodium oxygen and hydrogen atoms present.
NA=23 O=16 H=1 so we add 23+16+1=40 so 40 is our molecular relative mass
now we fix it in our formulae which is m=n× mr
m=1.5× 40 =60 so our mass is 60grams or 60g
HOPE THIS HELPS!!!! if i made a mistake our MAY answer may be wrong feel free to comment
The # 4 is the only significant # in 400.. Trailing 0's r not significant unless there is a decimal.
in 0.7000, there are 4 significant #'s.
Answer:
Two non bonded electron pairs and four bonded electron pairs
Explanation:
An image of the compound as obtained from chemlibretext is attached to this answer.
The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.
